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
• B. $4.0$ A
• C. $5.6$ A
• D. $8.0$ A

Answer 1

The correct option is C $5.6$ A

The relation between the rms current and the peak current is as follows:

$I_0=I_{rms}\sqrt{2}$

The expression for the heat produce by a resistor is given as:

$W=I^{2}Rt$

so, the heat produce by the resister when the steady current of $2A$ flow through it is,

$W_{2A}=\left(2A{)}^{2}R\right(1sec)$

$W_{2A}=4R$

now since the alternating current produces 4 time the heat produced by the steady current so, the heat produce by the alternating current is,

$W_{alternating}=4\times W_{2A}$

$W_{alternating}=16R$

$W_{alternating}=4^{2}R$

so, the rms current of the alternating current is $4A$

Now the peak current is as follows:

$I_0=\left(4A\right)\times \sqrt{2}$

$I_0=5.6A$