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  4. An alternating voltage given by V = 140 sin314t is connected across a pure resistor of 50 Ω, the rms current through the resistor is

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Q. An alternating voltage given by $V = 140\,sin314t$ is connected across a pure resistor of $50\,\Omega$, the rms current through the resistor is

Alternating Current

Solution:

Here $R=50\,\Omega$,
$V_{0}=140\,V$
$\therefore I_{rms}=\frac{V_{rms}}{R}$
$=\frac{0.707\,V_0}{R}$
$=\frac{0.707 \times 140}{50}$
$=1.98\,A$.