Question

A sinusoidal A.C current flows through the resistor of the resistance $R$. If the peak current is ${I_P}$, then power dissipated is,
A. ${I_P}^2R\cos \theta$
B. $\dfrac{1}{2}{I_P}^2R$
C. $4\pi {I_P}^2R$
D. $\dfrac{1}{{{\pi ^2}}}{I_P}^2R$

Answer 1

The voltage across the resistor and the current flowing through the resistor are both given in the sinusoidal quantities. The value of the total reduction in the circuit is important and the current value can be taken as the average current value. Use the formulas of power and average AC to solve the problem.
Formula used:
The formula for the power dissipated:
$\Rightarrow P = {I^2}R$
Where, $P$ is the power, the current is $I$ , and $R$ is the resistance.
The average A.C current value:
$\Rightarrow \dfrac{1}{{\sqrt 2 }} \times {I_P}$
Where, ${I_P}$is the peak current.

Complete step by step answer:
In the question, it is given that a sinusoidal A.C current flows through the resistor of the resistance.
The value of the peak current is given as ${I_P}$.
First, we can calculate the average current value. The average A.C current value is given as,
$\Rightarrow \dfrac{1}{{\sqrt 2 }} \times {I_P}$
It can also be written as,
$\Rightarrow \dfrac{{{I_P}}}{{\sqrt 2 }}$
The measure of the total reduction in the power that occurs because of the resistance of the circuit system is known as power dissipation. The power dissipation formula is given below:
$\Rightarrow P = {I^2}R$
Where, $P$ is the power, $I$ is current, and $R$ is the resistance.
In the question, they have asked us to calculate the value of the power dissipated when the peak current is ${I_P}$.
We have the value of the average current that is $\dfrac{{{I_P}}}{{\sqrt 2 }}$.
The r.m.s value of the current is ${I_{\max }} = \dfrac{{{I_P}}}{{\sqrt 2 }}$. Substitute the value in the power dissipation formula.
$\Rightarrow P = {I^2}R$
Consider the current value as the maximum current value and substitute in the formula.
$\Rightarrow P = \dfrac{{{I_P}}}{{\sqrt 2 }}R$
The equation can also be written as,
$\Rightarrow P = \dfrac{1}{{\sqrt 2 }}{I_P}R$
$\Rightarrow P = \dfrac{1}{2}{I_P}^2R$
Therefore, the value of the power dissipated when the peak current ${I_P}$ is $\dfrac{1}{2}{I_P}^2R$

So, the correct answer is “Option B”.

Note:
In a circuit, the current and voltage reach the minimum values and maximum values at the same time. The voltage across the resistor and the current flowing through the resistor are in phase with each other. The value of the root means the square of the alternating current is the steady current that will generate the same amount of heat in the circuit’s resistor.