A voltage \( V=5\cos(15\pi t)\,\mathrm{V} \) is applied across a resistor of \( 33\,\Omega \). What is the peak value of the current through the resistor?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.15 A
Across a resistor the current is in phase with the voltage, and its peak value is \( i_m=v_m/R \), whatever the frequency. Comparing \( V=5\cos(15\pi t)\,\mathrm{V} \) with \( V=v_m\cos\omega t \), the peak voltage is \( v_m=5\,\mathrm{V} \).
\( \begin{aligned} i_m &= \frac{v_m}{R} \\ &= \frac{5\,\mathrm{V}}{33\,\Omega} \\ &\approx 0.15\,\mathrm{A} \end{aligned} \)