If a 0.1 µF, 0.1 µF and 0.05 µF capacitor are connected in series across a 75 V source, the voltage drop across the smallest capacitor equals .
A)37.5 V
B)18.8 V
C)50 V
D)100 V

Question

Quizplus · Accepted Answer

Voltage drop across a capacitor is given by V = Q/C. Since the capacitors are connected in series, the charge on each capacitor is the same. Therefore, the voltage drop across the smallest capacitor (0.05 µF) is the highest.V = Q/C = (75 V) / (0.05 µF) = 37.5 V

If a 01 µF, 0