Question: Q. 4. An alternating voltage given by $V=140$ sin
$314 t$ is connected across a pure resistor of $50 \Omega$. Find :
(i) the frequency of the source.
(ii) the rms current through the resistor.
A [O.D. I, II, III 2012]
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Solution:
Ans. (i)
$$ \begin{aligned} 2 \pi f & =\omega \ 2 \pi f & =314 \mathrm{rad} \ f & =50 \mathrm{~Hz} \end{aligned} $$
$$ \Rightarrow \quad 2 \pi f=314 \mathrm{rad} \mathrm{s}^{-1} $$
(ii)
$$ \begin{aligned} & I_{r m s}=\frac{V_{r m s}}{R} \text { where, } V_{r m s}=\frac{V_{0}}{\sqrt{2}} \ & I_{r m s}=\frac{140}{\sqrt{2} \times 50}=1.98 \mathrm{~A} \approx 2 \mathrm{~A} \end{aligned} $$
[CBSE Marking Scheme 2012]
