Question

In an a.c. circuit with phase voltage $V$ and current $I$, the power dissipated is:
A. $\dfrac{{VI}}{2}$
B. $\dfrac{{VI}}{{\sqrt 2 }}$
C. $VI$
D. $VI\cos \theta$

Answer 1

To solve this question, we will consider RLC a.c. circuit whose total impedance is Z. We will first use the relation between power, voltage, current and resistance. After that, we will use the total impedance to find the required answer.In capacitive circuit, the circuit contains only capacitor and hence the phase difference between voltage and current is ${90^0}$ which means that $\theta = \dfrac{\pi }{2} \Rightarrow \cos \theta = 0$.Thus, there is no power dissipation in the capacitive circuit.

Formulas used:
$P = {I^2}R$ ,
where, $P$ is the dissipated power dissipated, $I$ is current and $R$ is resistance.
$I = \dfrac{V}{Z}$,
where, $I$ is current, $V$is voltage and $Z$ is total impedance of the circuit.
$\dfrac{R}{Z} = \cos \theta$,
where, $R$ is resistance, $Z$ is total impedance of the circuit and $\cos \theta$ is the power factor where $\theta$ is the phase difference between voltage and current.

Complete step by step answer:
We know that power dissipated is given by:
$P = {I^2}R$
We can rewrite this equation as
$P = IRI$
Now, we will put $I = \dfrac{V}{Z}$
$P = IR\dfrac{V}{Z}$
This can be rewritten as
$P = IV\dfrac{R}{Z}$
But, we know that $\dfrac{R}{Z} = \cos \theta$
$\therefore P = IV\cos \theta$
Thus power dissipated in a.c. circuit with phase voltage $V$and current $I$is $VI\cos \theta$.

Hence, option D is the right answer.

Note: Here, we have seen that power is dependent on values of voltage and current as well as the power factor. There are two important cases: resistive circuit and capacitive circuit. In a resistive circuit, the circuit contains only pure resistance and hence there is no phase difference between voltage and current which means that $\theta = 0 \Rightarrow \cos \theta = 1$. Thus, there is maximum power dissipation in a resistive circuit.