Question

The instantaneous current and voltage of an AC circuit are given by I = 10 sin 300tA and V= 200 sin 300t V. What is the power dissipation in the circuit?

Answer 1

- Hint: In order to solve this problem we need to get the root mean squared value of the current and voltage and then apply the formula of power to get the answer to this problem.

Complete step-by-step solution -

The given equations are I = 10 sin 300tA, V= 200 sin 300tV
The above equations are in the format $I = {I_0}\sin {\theta _1}$ and $V = {V_0}\sin {\theta _2}$
Where ${I_{0,\,}}{V_0}\,$are the amplitudes of current and voltages and ${\theta _1},\,{\theta _2}$ are the phase angles of current and voltages.
We know that the power = $P = {I_{rms}}{V_{rms}}\cos \phi$
And root mean squared value (RMS value) of the current and voltage is ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }} = \dfrac{{10}}{{\sqrt 2 }}$ and ${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }} = \dfrac{{200}}{{\sqrt 2 }}$.
The angle $\phi$ is the phase difference between current and voltage that is,
$\phi = {\theta _2} - {\theta _1}$= (300t – 300t) = 0
So, on putting the value of ${I_{rms}}$, ${V_{rms}}$ and $\phi$ in the formula of power we get,
Power = P = $\dfrac{{10}}{{\sqrt 2 }}{\text{x}}\dfrac{{200}}{{\sqrt 2 }}\cos 0$= $\dfrac{{2000}}{2} = 1000$
Hence, the power is 1000 watt.

Note – To solve this problem we just need to know some of the concepts those are the power = $P = {I_{rms}}{V_{rms}}\cos \phi$, the angle $\phi$ is the phase difference between current and voltage and ${I_{rms}}$, ${V_{rms}}$ are root mean squared current and voltage value ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$ and ${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$. The term cos$\phi$ is called the power factor or quality factor in alternating current. Knowing these formulas you will get the right answer to this problem.