A 315-$\varOmega$ resistor is connected in series with a parallel-plate capacitor across the terminals of a 60.0-Hz ac generator. When the gap between the plates is empty, its capacitance is 2.65 µF. Determine the ratio of the rms current in the circuit when the capacitor is empty to that when ruby mica ($\kappa$ = 5.40) is inserted between the plates?
A)2.87
B)2.32
C)0.430
D)0.348
E)0.253

Question

Quizplus · Accepted Answer

The impedance of the series connection of the resistor and capacitor is given by:
Z = R + Xc
where R is the resistance, Xc is the capacitive reactance which is given by Xc = 1/(2πfC), f is the frequency and C is the capacitance.

When the capacitor is empty: 
Xc1 = 1/(2π(60)(2.65x10^-6)) = 1.13x10^3 Ω
The impedance becomes: 
Z1 = 1.13x10^3 + 315 = 1.45x10^3Ω

When the ruby mica is inserted between the plates: 
Xc2 = 1/(2π(60)(5.40x10^-12)) = 4.14x10^3 Ω

The impedance becomes: 
Z2 = 4.14x10^3 + 315 = 4.45x10^3Ω

The ratio of rms current in both cases is given by: 
I2/I1 = Z1/Z2

Substitute the values of Z1 and Z2 and we get: 
I2/I1 = 1.45x10^3/4.45x10^3 = 0.326

Therefore, the ratio of the rms current in the circuit when the capacitor is empty to that when ruby mica is inserted between the plates is 0.326, which is closest to choice D (0.348).

A 315- Ω\varOmegaΩ Resistor Is Connected in Series with a Parallel-Plate Capacitor Across

A 315- $\varOmega$ Resistor Is Connected in Series with a Parallel-Plate Capacitor Across