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Question

What is the approximate peak value of an alternating current producing four times the heat produced per seconds by a steady current of 2 A in a resistor?

A
2.8 A
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B
4.0 A
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C
5.6 A
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D
8.0 A
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Solution

The correct option is C 5.6 A
The relation between the rms current and the peak current is as follows:
I0=Irms2
The expression for the heat produce by a resistor is given as:
W=I2Rt
so, the heat produce by the resister when the steady current of 2A flow through it is,
W2A=(2 A)2R(1 sec)
W2A=4R
now since the alternating current produces 4 time the heat produced by the steady current so, the heat produce by the alternating current is,
Walternating=4×W2A
Walternating=16R
Walternating=42R
so, the rms current of the alternating current is 4 A
Now the peak current is as follows:
I0=(4 A)×2
I0=5.6A

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