Problem 32

  Review Conceptual Example 5. A light bulb has a resistance of 240 $\Omega$. It is connected to a standard wall socket (120 V, 60.0 Hz). (a) Determine the current in the bulb. (b) Determine the current in the bulb after a 10.0-$\mu$F capacitor is added in series in the circuit. (c) It is possible to return the current in the bulb to the value calculated in part (a) by adding an inductor in series with the bulb and the capacitor. What is the value of the inductance of the inductor?

SOLUTION:
(a) Ohm's Law sayeth

$$I = \frac{V}{R} = 0.5 A .$$

(b) There's no inductor, so X_L = 0. The current is given by

$$I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}} ,$$

where the impedance of the capacitor is

$$X_C = \frac{1}{2 \pi f C} .$$

Hence, the current that floweth is

$$I = \frac{120V}{(240 \Omega)^2 + \left [ \frac{1}{2 \pi (60 Hz) (10 \times 10^{-6} F)} \right ]^2}$$

I = 0.34 A

(c) In part (a) we only had a resistor, so now we want to add an inductor in order to cancel the capacitor's impedance. Hence, in the formula for impedance,

$$Z = \sqrt{R^2 + (X_L - X_C)^2} ,$$

we want to have

X_L = X_C

$$2 \pi f L = \frac{1}{2 \pi f C}$$

$$L = \frac{1}{4 \pi^2 (60 Hz) (10 \times 10^{-6} F)}$$

L = 0.704 H .