Question

The voltage of an $AC$ source varies with time according to the equation $V=100\sin \left(100\pi t\right)\cos \left(100\pi t\right)$ . If this voltage is applied across a resistor of $25\Omega$, then,

• A. Current flowing in the circuit is $I=2\sin 200\pi t$.
• B. $rms$ value of current is $I_{rms}=\sqrt{2}\text{A}$.
• C. Current and voltage are in phase.
• D. $rms$ value of current is $I_{rms}=2\text{A}$.

Answer 1

Answer: (A), (B), (C)

Current and voltage are in phase.

Given:

$V=100\sin \left(100\pi t\right)\cos \left(100\pi t\right)$

$R=25\Omega$

Now,

$V=100\sin \left(100\pi t\right)\cos \left(100\pi t\right)$

$=50\times \left(2\sin 100\pi t\cos 100\pi t\right)$

$\therefore V=50\sin 200\pi t...\left(i\right)[\because 2\sin \theta \cos \theta =\sin 2\theta ]$

The current is given by,

$I=\frac{V}{R}$

$I=\frac{50\sin 200\pi t}{25}=2\sin \left(200\pi t\right).....\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, it is clear that voltage and current are in phase.

$rms$ value of current is,

$I_{rms}=\frac{2}{\sqrt{2}}=\sqrt{2}\text{A}$

Hence, options $\left(A\right),\left(B\right)$ and $\left(C\right)$ are correct.