Why do we use a transformer on some heating element projects when the wattage is the same on both sides of...
$begingroup$
On some projects involving heating elements there's a transformer being used.
and the explanation given is "we are using a transformer because need more current"
but since the watts on both sides of the Transformer are supposed to be identical,
wouldn't the heating element dissipate the same energy with and without the Transformer ?
power transformer heat
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
$endgroup$
|
show 5 more comments
$begingroup$
On some projects involving heating elements there's a transformer being used.
and the explanation given is "we are using a transformer because need more current"
but since the watts on both sides of the Transformer are supposed to be identical,
wouldn't the heating element dissipate the same energy with and without the Transformer ?
power transformer heat
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
$endgroup$
3
$begingroup$
Why are you talking about "wattage" when citing some current reasoning?
$endgroup$
– Eugene Sh.
Jan 14 at 22:53
$begingroup$
@EugeneSh. because i don't understand the current reasoning, i thought power is the same according to P=IV
$endgroup$
– soundslikefiziks
Jan 14 at 22:54
$begingroup$
The power is the same, but the voltage and the current can vary. If your heater is requiring more (or less) voltage than provided, you need a transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 22:55
1
$begingroup$
The resistor will dissipate $V^2/R$. Where $V$ will depend on the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:00
1
$begingroup$
Same as what? If the voltage on your supply is much higher than the $V$ above, then if you connect resistor directly to it the power will be much higher as well. The power (on both sides) is determined by the load, not by the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:07
|
show 5 more comments
$begingroup$
On some projects involving heating elements there's a transformer being used.
and the explanation given is "we are using a transformer because need more current"
but since the watts on both sides of the Transformer are supposed to be identical,
wouldn't the heating element dissipate the same energy with and without the Transformer ?
power transformer heat
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
$endgroup$
On some projects involving heating elements there's a transformer being used.
and the explanation given is "we are using a transformer because need more current"
but since the watts on both sides of the Transformer are supposed to be identical,
wouldn't the heating element dissipate the same energy with and without the Transformer ?
power transformer heat
power transformer heat
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
asked Jan 14 at 22:49
soundslikefizikssoundslikefiziks
486317
486317
3
$begingroup$
Why are you talking about "wattage" when citing some current reasoning?
$endgroup$
– Eugene Sh.
Jan 14 at 22:53
$begingroup$
@EugeneSh. because i don't understand the current reasoning, i thought power is the same according to P=IV
$endgroup$
– soundslikefiziks
Jan 14 at 22:54
$begingroup$
The power is the same, but the voltage and the current can vary. If your heater is requiring more (or less) voltage than provided, you need a transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 22:55
1
$begingroup$
The resistor will dissipate $V^2/R$. Where $V$ will depend on the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:00
1
$begingroup$
Same as what? If the voltage on your supply is much higher than the $V$ above, then if you connect resistor directly to it the power will be much higher as well. The power (on both sides) is determined by the load, not by the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:07
|
show 5 more comments
3
$begingroup$
Why are you talking about "wattage" when citing some current reasoning?
$endgroup$
– Eugene Sh.
Jan 14 at 22:53
$begingroup$
@EugeneSh. because i don't understand the current reasoning, i thought power is the same according to P=IV
$endgroup$
– soundslikefiziks
Jan 14 at 22:54
$begingroup$
The power is the same, but the voltage and the current can vary. If your heater is requiring more (or less) voltage than provided, you need a transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 22:55
1
$begingroup$
The resistor will dissipate $V^2/R$. Where $V$ will depend on the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:00
1
$begingroup$
Same as what? If the voltage on your supply is much higher than the $V$ above, then if you connect resistor directly to it the power will be much higher as well. The power (on both sides) is determined by the load, not by the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:07
3
3
$begingroup$
Why are you talking about "wattage" when citing some current reasoning?
$endgroup$
– Eugene Sh.
Jan 14 at 22:53
$begingroup$
Why are you talking about "wattage" when citing some current reasoning?
$endgroup$
– Eugene Sh.
Jan 14 at 22:53
$begingroup$
@EugeneSh. because i don't understand the current reasoning, i thought power is the same according to P=IV
$endgroup$
– soundslikefiziks
Jan 14 at 22:54
$begingroup$
@EugeneSh. because i don't understand the current reasoning, i thought power is the same according to P=IV
$endgroup$
– soundslikefiziks
Jan 14 at 22:54
$begingroup$
The power is the same, but the voltage and the current can vary. If your heater is requiring more (or less) voltage than provided, you need a transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 22:55
$begingroup$
The power is the same, but the voltage and the current can vary. If your heater is requiring more (or less) voltage than provided, you need a transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 22:55
1
1
$begingroup$
The resistor will dissipate $V^2/R$. Where $V$ will depend on the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:00
$begingroup$
The resistor will dissipate $V^2/R$. Where $V$ will depend on the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:00
1
1
$begingroup$
Same as what? If the voltage on your supply is much higher than the $V$ above, then if you connect resistor directly to it the power will be much higher as well. The power (on both sides) is determined by the load, not by the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:07
$begingroup$
Same as what? If the voltage on your supply is much higher than the $V$ above, then if you connect resistor directly to it the power will be much higher as well. The power (on both sides) is determined by the load, not by the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:07
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Power in a heater is given by the equation $ P = VI $ where P is power (watts), V the applied voltage (volts) and I the current (amps). You can see that there are multiple ways to achieve a given power - high voltage / low current - low voltage / high current.
For a transformer the relationship between input and output is given by $ P_{in} = P_{out} $ (ignoring the few percent losses in the transformer). From the previous formula we can write $ V_{in}I_{in} = V_{out}I_{out} $. This allows us to step the voltage up or down to meet the requirements of the load.
... but since the watts on both sides of the transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the transformer?
Yes, if the resistance of the wire is the same. A high voltage heater will use a very thin wire with high resistance. If this proves too fragile or too thin for the application then a thicker wire can be used but the current will have to increase to get the same current density in the wire. Since the wire is thicker it has a lower resistance so a lower voltage can be used too. An example is the butcher's bag sealing machine which uses an 'impulse' element nichrome (or similar) wire to seal the bag. Here a certain thickness of seal may be required and so a lower voltage, higher current is required. The transformer 'transforms' the voltage and current to match the heater and provides electrical isolation from the mains eliminating a shock hazard.
answered Jan 14 at 23:17
TransistorTransistor
82k778176
$endgroup$
$begingroup$
"and provides electrical isolation from the mains eliminating a shock hazard." so the benefit of connecting a transformer after a household socket is just so that the high current wont switch on the breaker ?
$endgroup$
– soundslikefiziks
Jan 14 at 23:30
$begingroup$
i mean, assuming we have a thicker cable like you say and we need to run more current through it, we can theoretically connect it to the 110v outlet without a breaker, and it could potentially provide much more current than it's stepped down voltage on the transformer. so "bypassing the breaker amp limit" is the only benefit i see now for using a transformer. (but i am sure my understanding still hasn't fully clicked in. so that can't be the only usage for a transformer)
$endgroup$
– soundslikefiziks
Jan 14 at 23:43
1
$begingroup$
Since the resistance is fixed, to set the power you need to vary the voltage and current. This is what the transformer is for.
$endgroup$
– evildemonic
Jan 14 at 23:53
1
$begingroup$
@soundslikefiziks I think you've missed two points: 1) Stepping down the voltage makes it safer - if the voltage is less than 50V you are unlikely to be electrocuted if you touch the element. 2) For a given wattage, you can use high voltage and low current, or low voltage and high current. High voltage and low current means a high resistance - a long, thin resistance wire which may be fragile. Low voltage and high current implies a short, fat resistance wire, which is sturdier.
$endgroup$
– Simon B
Jan 15 at 0:09
1
$begingroup$
@soundslikefiziks - I think your main sticking point is thinking that wattage RATING is wattage consumed. The power supply can provide a specific wattage before it stops working within spec (usually by dropping the voltage). But that is MAXIMUM wattage. A heating element will consume wattage depending on supplied voltage - not wattage of the power supply. You are confusing MAXIMUM possible wattage supply with wattage in the circuit
$endgroup$
– slebetman
Jan 15 at 1:25
|
show 4 more comments
$begingroup$
If you have a 1000 Watt kettle designed for use in North America, where we have 120V power in the kitchen, then take it to Europe where they have 240V, the kettle will consume 4000 watts (resistance unchanged, but double voltage, power is voltage squared over resistance).
To use the kettle safely in Europe, you will need a step-down transformer to convert the 240V supply to the 120V that the kettle expects. With the step-down transformer the kettle will consume 1000 watts, and the transformer will draw 1000 watts from the 240V source.
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
$endgroup$
add a comment |
$begingroup$
A heater is essentially a resistor with some resistance $R$ given in Ohms ($Ω$). Suppose we have (for example) a $36 W$ heater designed for $12 V$. This heater would have a resistance of $4 Ω$. If $12 V$ is connected to the heater, a current of $I = V/R = 12V/4Ω = 3 A$ would flow (Ohm's law). This would dissipate a power of $P = Vtimes I = 12Vtimes3A = 36 W$.
but since the watts on both sides of the Transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the Transformer?
If we connect the (nominally $36W$) heater directly to the ($220 V$) mains (without a $220V$ to $12V$ transformer), then a current of $220V/4 Ω=55A$ would flow, causing a power dissipation of $220Vtimes55A=12100W$. This will destroy the heater and/or blow the fuse/breaker.
A $36W$ heater designed for $220V$ would need to have a resistance of $1344Ω$. If we were to connect this heater to $12V$ it would only dissipate $0.1W$.
The power dissipated by a heater depends on the voltage applied to it. If you connect a heater to a too high voltage it will heat up too much, if you connect a heater to a too low voltage it won't heat up enough. The transformer is needed to get the correct voltage.
A transformer increases the current while decreasing the voltage (or vice-versa). In the scenario where the $36W$ heater is connected to a $220V$ to $12V$ transformer, the heater draws $3A$ from the $12V$ supply (provided by the transformer), but the transformer, in turn, draws only $0.16A$ from the $220V$ mains (ignoring losses in the transformer). So the transformer "increases" the current, but this is only part of the story.
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Power in a heater is given by the equation $ P = VI $ where P is power (watts), V the applied voltage (volts) and I the current (amps). You can see that there are multiple ways to achieve a given power - high voltage / low current - low voltage / high current.
For a transformer the relationship between input and output is given by $ P_{in} = P_{out} $ (ignoring the few percent losses in the transformer). From the previous formula we can write $ V_{in}I_{in} = V_{out}I_{out} $. This allows us to step the voltage up or down to meet the requirements of the load.
... but since the watts on both sides of the transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the transformer?
Yes, if the resistance of the wire is the same. A high voltage heater will use a very thin wire with high resistance. If this proves too fragile or too thin for the application then a thicker wire can be used but the current will have to increase to get the same current density in the wire. Since the wire is thicker it has a lower resistance so a lower voltage can be used too. An example is the butcher's bag sealing machine which uses an 'impulse' element nichrome (or similar) wire to seal the bag. Here a certain thickness of seal may be required and so a lower voltage, higher current is required. The transformer 'transforms' the voltage and current to match the heater and provides electrical isolation from the mains eliminating a shock hazard.
answered Jan 14 at 23:17
TransistorTransistor
82k778176
$endgroup$
$begingroup$
"and provides electrical isolation from the mains eliminating a shock hazard." so the benefit of connecting a transformer after a household socket is just so that the high current wont switch on the breaker ?
$endgroup$
– soundslikefiziks
Jan 14 at 23:30
$begingroup$
i mean, assuming we have a thicker cable like you say and we need to run more current through it, we can theoretically connect it to the 110v outlet without a breaker, and it could potentially provide much more current than it's stepped down voltage on the transformer. so "bypassing the breaker amp limit" is the only benefit i see now for using a transformer. (but i am sure my understanding still hasn't fully clicked in. so that can't be the only usage for a transformer)
$endgroup$
– soundslikefiziks
Jan 14 at 23:43
1
$begingroup$
Since the resistance is fixed, to set the power you need to vary the voltage and current. This is what the transformer is for.
$endgroup$
– evildemonic
Jan 14 at 23:53
1
$begingroup$
@soundslikefiziks I think you've missed two points: 1) Stepping down the voltage makes it safer - if the voltage is less than 50V you are unlikely to be electrocuted if you touch the element. 2) For a given wattage, you can use high voltage and low current, or low voltage and high current. High voltage and low current means a high resistance - a long, thin resistance wire which may be fragile. Low voltage and high current implies a short, fat resistance wire, which is sturdier.
$endgroup$
– Simon B
Jan 15 at 0:09
1
$begingroup$
@soundslikefiziks - I think your main sticking point is thinking that wattage RATING is wattage consumed. The power supply can provide a specific wattage before it stops working within spec (usually by dropping the voltage). But that is MAXIMUM wattage. A heating element will consume wattage depending on supplied voltage - not wattage of the power supply. You are confusing MAXIMUM possible wattage supply with wattage in the circuit
$endgroup$
– slebetman
Jan 15 at 1:25
|
show 4 more comments
$begingroup$
Power in a heater is given by the equation $ P = VI $ where P is power (watts), V the applied voltage (volts) and I the current (amps). You can see that there are multiple ways to achieve a given power - high voltage / low current - low voltage / high current.
For a transformer the relationship between input and output is given by $ P_{in} = P_{out} $ (ignoring the few percent losses in the transformer). From the previous formula we can write $ V_{in}I_{in} = V_{out}I_{out} $. This allows us to step the voltage up or down to meet the requirements of the load.
... but since the watts on both sides of the transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the transformer?
Yes, if the resistance of the wire is the same. A high voltage heater will use a very thin wire with high resistance. If this proves too fragile or too thin for the application then a thicker wire can be used but the current will have to increase to get the same current density in the wire. Since the wire is thicker it has a lower resistance so a lower voltage can be used too. An example is the butcher's bag sealing machine which uses an 'impulse' element nichrome (or similar) wire to seal the bag. Here a certain thickness of seal may be required and so a lower voltage, higher current is required. The transformer 'transforms' the voltage and current to match the heater and provides electrical isolation from the mains eliminating a shock hazard.
answered Jan 14 at 23:17
TransistorTransistor
82k778176
$endgroup$
$begingroup$
"and provides electrical isolation from the mains eliminating a shock hazard." so the benefit of connecting a transformer after a household socket is just so that the high current wont switch on the breaker ?
$endgroup$
– soundslikefiziks
Jan 14 at 23:30
$begingroup$
i mean, assuming we have a thicker cable like you say and we need to run more current through it, we can theoretically connect it to the 110v outlet without a breaker, and it could potentially provide much more current than it's stepped down voltage on the transformer. so "bypassing the breaker amp limit" is the only benefit i see now for using a transformer. (but i am sure my understanding still hasn't fully clicked in. so that can't be the only usage for a transformer)
$endgroup$
– soundslikefiziks
Jan 14 at 23:43
1
$begingroup$
Since the resistance is fixed, to set the power you need to vary the voltage and current. This is what the transformer is for.
$endgroup$
– evildemonic
Jan 14 at 23:53
1
$begingroup$
@soundslikefiziks I think you've missed two points: 1) Stepping down the voltage makes it safer - if the voltage is less than 50V you are unlikely to be electrocuted if you touch the element. 2) For a given wattage, you can use high voltage and low current, or low voltage and high current. High voltage and low current means a high resistance - a long, thin resistance wire which may be fragile. Low voltage and high current implies a short, fat resistance wire, which is sturdier.
$endgroup$
– Simon B
Jan 15 at 0:09
1
$begingroup$
@soundslikefiziks - I think your main sticking point is thinking that wattage RATING is wattage consumed. The power supply can provide a specific wattage before it stops working within spec (usually by dropping the voltage). But that is MAXIMUM wattage. A heating element will consume wattage depending on supplied voltage - not wattage of the power supply. You are confusing MAXIMUM possible wattage supply with wattage in the circuit
$endgroup$
– slebetman
Jan 15 at 1:25
|
show 4 more comments
$begingroup$
Power in a heater is given by the equation $ P = VI $ where P is power (watts), V the applied voltage (volts) and I the current (amps). You can see that there are multiple ways to achieve a given power - high voltage / low current - low voltage / high current.
For a transformer the relationship between input and output is given by $ P_{in} = P_{out} $ (ignoring the few percent losses in the transformer). From the previous formula we can write $ V_{in}I_{in} = V_{out}I_{out} $. This allows us to step the voltage up or down to meet the requirements of the load.
... but since the watts on both sides of the transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the transformer?
Yes, if the resistance of the wire is the same. A high voltage heater will use a very thin wire with high resistance. If this proves too fragile or too thin for the application then a thicker wire can be used but the current will have to increase to get the same current density in the wire. Since the wire is thicker it has a lower resistance so a lower voltage can be used too. An example is the butcher's bag sealing machine which uses an 'impulse' element nichrome (or similar) wire to seal the bag. Here a certain thickness of seal may be required and so a lower voltage, higher current is required. The transformer 'transforms' the voltage and current to match the heater and provides electrical isolation from the mains eliminating a shock hazard.
answered Jan 14 at 23:17
TransistorTransistor
82k778176
$endgroup$
Power in a heater is given by the equation $ P = VI $ where P is power (watts), V the applied voltage (volts) and I the current (amps). You can see that there are multiple ways to achieve a given power - high voltage / low current - low voltage / high current.
For a transformer the relationship between input and output is given by $ P_{in} = P_{out} $ (ignoring the few percent losses in the transformer). From the previous formula we can write $ V_{in}I_{in} = V_{out}I_{out} $. This allows us to step the voltage up or down to meet the requirements of the load.
... but since the watts on both sides of the transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the transformer?
Yes, if the resistance of the wire is the same. A high voltage heater will use a very thin wire with high resistance. If this proves too fragile or too thin for the application then a thicker wire can be used but the current will have to increase to get the same current density in the wire. Since the wire is thicker it has a lower resistance so a lower voltage can be used too. An example is the butcher's bag sealing machine which uses an 'impulse' element nichrome (or similar) wire to seal the bag. Here a certain thickness of seal may be required and so a lower voltage, higher current is required. The transformer 'transforms' the voltage and current to match the heater and provides electrical isolation from the mains eliminating a shock hazard.
answered Jan 14 at 23:17
TransistorTransistor
82k778176
answered Jan 14 at 23:17
TransistorTransistor
82k778176
answered Jan 14 at 23:17
TransistorTransistor
82k778176
answered Jan 14 at 23:17
TransistorTransistor
82k778176
82k778176
$begingroup$
"and provides electrical isolation from the mains eliminating a shock hazard." so the benefit of connecting a transformer after a household socket is just so that the high current wont switch on the breaker ?
$endgroup$
– soundslikefiziks
Jan 14 at 23:30
$begingroup$
i mean, assuming we have a thicker cable like you say and we need to run more current through it, we can theoretically connect it to the 110v outlet without a breaker, and it could potentially provide much more current than it's stepped down voltage on the transformer. so "bypassing the breaker amp limit" is the only benefit i see now for using a transformer. (but i am sure my understanding still hasn't fully clicked in. so that can't be the only usage for a transformer)
$endgroup$
– soundslikefiziks
Jan 14 at 23:43
1
$begingroup$
Since the resistance is fixed, to set the power you need to vary the voltage and current. This is what the transformer is for.
$endgroup$
– evildemonic
Jan 14 at 23:53
1
$begingroup$
@soundslikefiziks I think you've missed two points: 1) Stepping down the voltage makes it safer - if the voltage is less than 50V you are unlikely to be electrocuted if you touch the element. 2) For a given wattage, you can use high voltage and low current, or low voltage and high current. High voltage and low current means a high resistance - a long, thin resistance wire which may be fragile. Low voltage and high current implies a short, fat resistance wire, which is sturdier.
$endgroup$
– Simon B
Jan 15 at 0:09
1
$begingroup$
@soundslikefiziks - I think your main sticking point is thinking that wattage RATING is wattage consumed. The power supply can provide a specific wattage before it stops working within spec (usually by dropping the voltage). But that is MAXIMUM wattage. A heating element will consume wattage depending on supplied voltage - not wattage of the power supply. You are confusing MAXIMUM possible wattage supply with wattage in the circuit
$endgroup$
– slebetman
Jan 15 at 1:25
|
show 4 more comments
$begingroup$
"and provides electrical isolation from the mains eliminating a shock hazard." so the benefit of connecting a transformer after a household socket is just so that the high current wont switch on the breaker ?
$endgroup$
– soundslikefiziks
Jan 14 at 23:30
$begingroup$
i mean, assuming we have a thicker cable like you say and we need to run more current through it, we can theoretically connect it to the 110v outlet without a breaker, and it could potentially provide much more current than it's stepped down voltage on the transformer. so "bypassing the breaker amp limit" is the only benefit i see now for using a transformer. (but i am sure my understanding still hasn't fully clicked in. so that can't be the only usage for a transformer)
$endgroup$
– soundslikefiziks
Jan 14 at 23:43
1
$begingroup$
Since the resistance is fixed, to set the power you need to vary the voltage and current. This is what the transformer is for.
$endgroup$
– evildemonic
Jan 14 at 23:53
1
$begingroup$
@soundslikefiziks I think you've missed two points: 1) Stepping down the voltage makes it safer - if the voltage is less than 50V you are unlikely to be electrocuted if you touch the element. 2) For a given wattage, you can use high voltage and low current, or low voltage and high current. High voltage and low current means a high resistance - a long, thin resistance wire which may be fragile. Low voltage and high current implies a short, fat resistance wire, which is sturdier.
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– Simon B
Jan 15 at 0:09
1
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@soundslikefiziks - I think your main sticking point is thinking that wattage RATING is wattage consumed. The power supply can provide a specific wattage before it stops working within spec (usually by dropping the voltage). But that is MAXIMUM wattage. A heating element will consume wattage depending on supplied voltage - not wattage of the power supply. You are confusing MAXIMUM possible wattage supply with wattage in the circuit
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– slebetman
Jan 15 at 1:25
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"and provides electrical isolation from the mains eliminating a shock hazard." so the benefit of connecting a transformer after a household socket is just so that the high current wont switch on the breaker ?
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– soundslikefiziks
Jan 14 at 23:30
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"and provides electrical isolation from the mains eliminating a shock hazard." so the benefit of connecting a transformer after a household socket is just so that the high current wont switch on the breaker ?
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– soundslikefiziks
Jan 14 at 23:30
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i mean, assuming we have a thicker cable like you say and we need to run more current through it, we can theoretically connect it to the 110v outlet without a breaker, and it could potentially provide much more current than it's stepped down voltage on the transformer. so "bypassing the breaker amp limit" is the only benefit i see now for using a transformer. (but i am sure my understanding still hasn't fully clicked in. so that can't be the only usage for a transformer)
$endgroup$
– soundslikefiziks
Jan 14 at 23:43
$begingroup$
i mean, assuming we have a thicker cable like you say and we need to run more current through it, we can theoretically connect it to the 110v outlet without a breaker, and it could potentially provide much more current than it's stepped down voltage on the transformer. so "bypassing the breaker amp limit" is the only benefit i see now for using a transformer. (but i am sure my understanding still hasn't fully clicked in. so that can't be the only usage for a transformer)
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– soundslikefiziks
Jan 14 at 23:43
1
1
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Since the resistance is fixed, to set the power you need to vary the voltage and current. This is what the transformer is for.
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– evildemonic
Jan 14 at 23:53
$begingroup$
Since the resistance is fixed, to set the power you need to vary the voltage and current. This is what the transformer is for.
$endgroup$
– evildemonic
Jan 14 at 23:53
1
1
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@soundslikefiziks I think you've missed two points: 1) Stepping down the voltage makes it safer - if the voltage is less than 50V you are unlikely to be electrocuted if you touch the element. 2) For a given wattage, you can use high voltage and low current, or low voltage and high current. High voltage and low current means a high resistance - a long, thin resistance wire which may be fragile. Low voltage and high current implies a short, fat resistance wire, which is sturdier.
$endgroup$
– Simon B
Jan 15 at 0:09
$begingroup$
@soundslikefiziks I think you've missed two points: 1) Stepping down the voltage makes it safer - if the voltage is less than 50V you are unlikely to be electrocuted if you touch the element. 2) For a given wattage, you can use high voltage and low current, or low voltage and high current. High voltage and low current means a high resistance - a long, thin resistance wire which may be fragile. Low voltage and high current implies a short, fat resistance wire, which is sturdier.
$endgroup$
– Simon B
Jan 15 at 0:09
1
1
$begingroup$
@soundslikefiziks - I think your main sticking point is thinking that wattage RATING is wattage consumed. The power supply can provide a specific wattage before it stops working within spec (usually by dropping the voltage). But that is MAXIMUM wattage. A heating element will consume wattage depending on supplied voltage - not wattage of the power supply. You are confusing MAXIMUM possible wattage supply with wattage in the circuit
$endgroup$
– slebetman
Jan 15 at 1:25
$begingroup$
@soundslikefiziks - I think your main sticking point is thinking that wattage RATING is wattage consumed. The power supply can provide a specific wattage before it stops working within spec (usually by dropping the voltage). But that is MAXIMUM wattage. A heating element will consume wattage depending on supplied voltage - not wattage of the power supply. You are confusing MAXIMUM possible wattage supply with wattage in the circuit
$endgroup$
– slebetman
Jan 15 at 1:25
|
show 4 more comments
$begingroup$
If you have a 1000 Watt kettle designed for use in North America, where we have 120V power in the kitchen, then take it to Europe where they have 240V, the kettle will consume 4000 watts (resistance unchanged, but double voltage, power is voltage squared over resistance).
To use the kettle safely in Europe, you will need a step-down transformer to convert the 240V supply to the 120V that the kettle expects. With the step-down transformer the kettle will consume 1000 watts, and the transformer will draw 1000 watts from the 240V source.
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
$endgroup$
add a comment |
$begingroup$
If you have a 1000 Watt kettle designed for use in North America, where we have 120V power in the kitchen, then take it to Europe where they have 240V, the kettle will consume 4000 watts (resistance unchanged, but double voltage, power is voltage squared over resistance).
To use the kettle safely in Europe, you will need a step-down transformer to convert the 240V supply to the 120V that the kettle expects. With the step-down transformer the kettle will consume 1000 watts, and the transformer will draw 1000 watts from the 240V source.
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
$endgroup$
add a comment |
$begingroup$
If you have a 1000 Watt kettle designed for use in North America, where we have 120V power in the kitchen, then take it to Europe where they have 240V, the kettle will consume 4000 watts (resistance unchanged, but double voltage, power is voltage squared over resistance).
To use the kettle safely in Europe, you will need a step-down transformer to convert the 240V supply to the 120V that the kettle expects. With the step-down transformer the kettle will consume 1000 watts, and the transformer will draw 1000 watts from the 240V source.
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
$endgroup$
If you have a 1000 Watt kettle designed for use in North America, where we have 120V power in the kitchen, then take it to Europe where they have 240V, the kettle will consume 4000 watts (resistance unchanged, but double voltage, power is voltage squared over resistance).
To use the kettle safely in Europe, you will need a step-down transformer to convert the 240V supply to the 120V that the kettle expects. With the step-down transformer the kettle will consume 1000 watts, and the transformer will draw 1000 watts from the 240V source.
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
answered Jan 14 at 23:55
Peter BennettPeter Bennett
37.2k12967
37.2k12967
add a comment |
add a comment |
$begingroup$
A heater is essentially a resistor with some resistance $R$ given in Ohms ($Ω$). Suppose we have (for example) a $36 W$ heater designed for $12 V$. This heater would have a resistance of $4 Ω$. If $12 V$ is connected to the heater, a current of $I = V/R = 12V/4Ω = 3 A$ would flow (Ohm's law). This would dissipate a power of $P = Vtimes I = 12Vtimes3A = 36 W$.
but since the watts on both sides of the Transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the Transformer?
If we connect the (nominally $36W$) heater directly to the ($220 V$) mains (without a $220V$ to $12V$ transformer), then a current of $220V/4 Ω=55A$ would flow, causing a power dissipation of $220Vtimes55A=12100W$. This will destroy the heater and/or blow the fuse/breaker.
A $36W$ heater designed for $220V$ would need to have a resistance of $1344Ω$. If we were to connect this heater to $12V$ it would only dissipate $0.1W$.
The power dissipated by a heater depends on the voltage applied to it. If you connect a heater to a too high voltage it will heat up too much, if you connect a heater to a too low voltage it won't heat up enough. The transformer is needed to get the correct voltage.
A transformer increases the current while decreasing the voltage (or vice-versa). In the scenario where the $36W$ heater is connected to a $220V$ to $12V$ transformer, the heater draws $3A$ from the $12V$ supply (provided by the transformer), but the transformer, in turn, draws only $0.16A$ from the $220V$ mains (ignoring losses in the transformer). So the transformer "increases" the current, but this is only part of the story.
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
$endgroup$
add a comment |
$begingroup$
A heater is essentially a resistor with some resistance $R$ given in Ohms ($Ω$). Suppose we have (for example) a $36 W$ heater designed for $12 V$. This heater would have a resistance of $4 Ω$. If $12 V$ is connected to the heater, a current of $I = V/R = 12V/4Ω = 3 A$ would flow (Ohm's law). This would dissipate a power of $P = Vtimes I = 12Vtimes3A = 36 W$.
but since the watts on both sides of the Transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the Transformer?
If we connect the (nominally $36W$) heater directly to the ($220 V$) mains (without a $220V$ to $12V$ transformer), then a current of $220V/4 Ω=55A$ would flow, causing a power dissipation of $220Vtimes55A=12100W$. This will destroy the heater and/or blow the fuse/breaker.
A $36W$ heater designed for $220V$ would need to have a resistance of $1344Ω$. If we were to connect this heater to $12V$ it would only dissipate $0.1W$.
The power dissipated by a heater depends on the voltage applied to it. If you connect a heater to a too high voltage it will heat up too much, if you connect a heater to a too low voltage it won't heat up enough. The transformer is needed to get the correct voltage.
A transformer increases the current while decreasing the voltage (or vice-versa). In the scenario where the $36W$ heater is connected to a $220V$ to $12V$ transformer, the heater draws $3A$ from the $12V$ supply (provided by the transformer), but the transformer, in turn, draws only $0.16A$ from the $220V$ mains (ignoring losses in the transformer). So the transformer "increases" the current, but this is only part of the story.
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
$endgroup$
add a comment |
$begingroup$
A heater is essentially a resistor with some resistance $R$ given in Ohms ($Ω$). Suppose we have (for example) a $36 W$ heater designed for $12 V$. This heater would have a resistance of $4 Ω$. If $12 V$ is connected to the heater, a current of $I = V/R = 12V/4Ω = 3 A$ would flow (Ohm's law). This would dissipate a power of $P = Vtimes I = 12Vtimes3A = 36 W$.
but since the watts on both sides of the Transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the Transformer?
If we connect the (nominally $36W$) heater directly to the ($220 V$) mains (without a $220V$ to $12V$ transformer), then a current of $220V/4 Ω=55A$ would flow, causing a power dissipation of $220Vtimes55A=12100W$. This will destroy the heater and/or blow the fuse/breaker.
A $36W$ heater designed for $220V$ would need to have a resistance of $1344Ω$. If we were to connect this heater to $12V$ it would only dissipate $0.1W$.
The power dissipated by a heater depends on the voltage applied to it. If you connect a heater to a too high voltage it will heat up too much, if you connect a heater to a too low voltage it won't heat up enough. The transformer is needed to get the correct voltage.
A transformer increases the current while decreasing the voltage (or vice-versa). In the scenario where the $36W$ heater is connected to a $220V$ to $12V$ transformer, the heater draws $3A$ from the $12V$ supply (provided by the transformer), but the transformer, in turn, draws only $0.16A$ from the $220V$ mains (ignoring losses in the transformer). So the transformer "increases" the current, but this is only part of the story.
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
$endgroup$
A heater is essentially a resistor with some resistance $R$ given in Ohms ($Ω$). Suppose we have (for example) a $36 W$ heater designed for $12 V$. This heater would have a resistance of $4 Ω$. If $12 V$ is connected to the heater, a current of $I = V/R = 12V/4Ω = 3 A$ would flow (Ohm's law). This would dissipate a power of $P = Vtimes I = 12Vtimes3A = 36 W$.
but since the watts on both sides of the Transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the Transformer?
If we connect the (nominally $36W$) heater directly to the ($220 V$) mains (without a $220V$ to $12V$ transformer), then a current of $220V/4 Ω=55A$ would flow, causing a power dissipation of $220Vtimes55A=12100W$. This will destroy the heater and/or blow the fuse/breaker.
A $36W$ heater designed for $220V$ would need to have a resistance of $1344Ω$. If we were to connect this heater to $12V$ it would only dissipate $0.1W$.
The power dissipated by a heater depends on the voltage applied to it. If you connect a heater to a too high voltage it will heat up too much, if you connect a heater to a too low voltage it won't heat up enough. The transformer is needed to get the correct voltage.
A transformer increases the current while decreasing the voltage (or vice-versa). In the scenario where the $36W$ heater is connected to a $220V$ to $12V$ transformer, the heater draws $3A$ from the $12V$ supply (provided by the transformer), but the transformer, in turn, draws only $0.16A$ from the $220V$ mains (ignoring losses in the transformer). So the transformer "increases" the current, but this is only part of the story.
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
answered Jan 15 at 10:34
Tom van der ZandenTom van der Zanden
1513
1513
add a comment |
add a comment |
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3
$begingroup$
Why are you talking about "wattage" when citing some current reasoning?
$endgroup$
– Eugene Sh.
Jan 14 at 22:53
$begingroup$
@EugeneSh. because i don't understand the current reasoning, i thought power is the same according to P=IV
$endgroup$
– soundslikefiziks
Jan 14 at 22:54
$begingroup$
The power is the same, but the voltage and the current can vary. If your heater is requiring more (or less) voltage than provided, you need a transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 22:55
1
$begingroup$
The resistor will dissipate $V^2/R$. Where $V$ will depend on the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:00
1
$begingroup$
Same as what? If the voltage on your supply is much higher than the $V$ above, then if you connect resistor directly to it the power will be much higher as well. The power (on both sides) is determined by the load, not by the transformer.
$endgroup$
– Eugene Sh.
Jan 14 at 23:07