I have a soldering iron of 220v and 25w but it gets extremely reddish hot when i plug it. I calculated the resistance to reduce amperage using ohms law (R=V/I , R=220/1.5=147ohms approx) So my question is that whit is the wattage of resistor and how to calculate it? Is what i am doing is correct or not?
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Dear Dishendra Mishra has calculated out the current as 1.5 amp. But, has not given the work out; so it is not understood how it became 1.5 amp.
If the tolerance factors and AC peak factors are considered negligible and omitted from the calculation (as dear D.M. did) the rough RMS value of potential difference 220 volt divided by current in ampear should give out resistance in a passive component, alright. The wattage is calculated by potential difference multiplied by current consumed. So if wattage divided by potential difference will give out value of current. Here it would be :- 25W / 220V = 0.1137 A → this put in the V/I = R formula → R = 220V / 0.1137 A
→ 1934 Ω = 1.934 or 2 kΩ
By no means 1.5 A current could be worked out.
Required clarification or correction if it is mis-written; otherwise no positive help is countably possible. Thanks.
If the tolerance factors and AC peak factors are considered negligible and omitted from the calculation (as dear D.M. did) the rough RMS value of potential difference 220 volt divided by current in ampear should give out resistance in a passive component, alright. The wattage is calculated by potential difference multiplied by current consumed. So if wattage divided by potential difference will give out value of current. Here it would be :- 25W / 220V = 0.1137 A → this put in the V/I = R formula → R = 220V / 0.1137 A
→ 1934 Ω = 1.934 or 2 kΩ
By no means 1.5 A current could be worked out.
Required clarification or correction if it is mis-written; otherwise no positive help is countably possible. Thanks.
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You have asked a question and have answered it very well yourself. That is the way I would have answered.
You have rightly calculated the resistance of the soldering iron to be 147 ohms.You are right everywhere.
You have rightly calculated the resistance of the soldering iron to be 147 ohms.You are right everywhere.
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No, you made a mistake in calculating the current...
You know that:
P = U * I (Watt = Volt * Ampere)
so
I = P / U (Ampere = Watt / Volt)
I = 25 / 220 = 0.1136... A
You also know that:
U = R * I (Volt = Ohm * Ampere)
so
R = U / I
R = 220 / 0.1136 = 1936 Ohm
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R = U / I
and
I = P / U
so
R = U / (P / U)
or
R = U² / P (Ohm = Volt² / Watt)
R = 220² / 25 = 48400 / 25 = 1936 Ohm
You know that:
P = U * I (Watt = Volt * Ampere)
so
I = P / U (Ampere = Watt / Volt)
I = 25 / 220 = 0.1136... A
You also know that:
U = R * I (Volt = Ohm * Ampere)
so
R = U / I
R = 220 / 0.1136 = 1936 Ohm
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R = U / I
and
I = P / U
so
R = U / (P / U)
or
R = U² / P (Ohm = Volt² / Watt)
R = 220² / 25 = 48400 / 25 = 1936 Ohm
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You can find answers to such questions on conceptualphysicstoday.com
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here Voltage is given = 220V and wattage is given = 25 watt
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ok