Question

What is the approximate peak value of an alternating current producing four times the heat produced per second by a steady current of $2\;{\text{A}}$ in a resistor?
A. $2.8\;{\text{A}}$
B. $4.0\;{\text{A}}$
C. $5.6\;{\text{A}}$
D. $8.4\;{\text{A}}$

Answer 1

Hint : Root mean square value of current flowing through the circuit is twice the value of steady-state current flowing and the ratio of maximum current flowing through the circuit to the root mean square value of current flowing through the circuit is the square root of two. By using this data, the approximate peak value of alternate current flowing through the circuit can be calculated.

Useful formula:
The expression for finding the root mean square value of alternative current is
${I_{rms}} = 2 \times {I_s}$
Where ${I_{rms}}$ RMS value of alternate current and ${I_s}$is the value of steady state current.

${I_{\max }} = \sqrt 2 \;{I_{rms}}$

${I_{\max }}$is the peak value of current flowing through the circuit.

Given data:
Steady state current in a resistor is ${I_s} = 2\;{\text{A}}$

Complete step-by-step solution:

The expression for finding the root mean square value of alternative current is
${I_{rms}} = 2 \times {I_s}$
Substitute all the values in the above equation.
$

{I_{rms}} = 2 \times 2;{\text{A}} \\
{I_{rms}} = 4;{\text{A}} \\
$

The expression for finding the peak value of current flowing through the resistor is:
${I_{\max }} = \sqrt 2 \;{I_{rms}}$
Substitute all the values in the above equation.
$

{I_{\max }} = \sqrt 2 ; \times 4;A \\
{I_{\max }} = 1.414; \times 4;A \\
{I_{\max }} = 5.6;{\text{A}} \\
; \\
$

Thus, the peak value of current flowing through the resistor is ${I_{\max }} = 5.6\;{\text{A}}$

Hence the option C is correct.

Note : The thermal energy developed in the resistor when the sinusoidal current passes through the resistor depends on the root mean square value of the current, resistance, and the time period. The thermal energy is produced in the negative and positive intervals of the current flow.