An alternating voltage \( V = 140\sin 100\pi t \) is connected across a pure resistor of resistance \( 40\,\Omega \). What is the rms current through the resistor? (Take \( \sqrt{2} = 1.4 \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.5 A
In \( V = V_0\sin\omega t \) the coefficient is the peak voltage, \( V_0 = 140\,\mathrm{V} \). The rms value of a sinusoidal quantity is its peak value divided by \( \sqrt{2} \), and for a pure resistor Ohm's law holds for the rms values.
\( \displaystyle V_{\mathrm{rms}} = \frac{V_0}{\sqrt{2}} = \frac{140}{1.4} = 100\,\mathrm{V} \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{R} = \frac{100\,\mathrm{V}}{40\,\Omega} = 2.5\,\mathrm{A} \)