Question

$P26_75.jpg$

Some light dimmer switches use a variable resistor as shown in the figure above. The slide moves from position x = 0 to x = 1, and the resistance up to the slide position x is proportional to x (the total resistance R_pot = 100 Ω). If the resistance of the bulb is R_bulb = 397 Ω, what is the power expended in the lightbulb if

a) x = 1.00
P =

b) x = 0.50
P =

c) x = 0.25
P =

Answer 1

The variable resistor and

(a) x=1 R=100 the bulb is connected in parallel, therefore PD across each resistor is same

$P=\frac{V^2}{R}=\frac{120^2}{397} = 36.27 \ W$

(b) x=0.5 R=50

The effective series is we have R1=50 in series with parallel combination of R2=50 and Rbulb=397

$\\ R_{eq}=50+\frac{50\times 397}{397+50} = 94.41 \ \Omega \\ \text{ effective current in the circuit} \\ I=\frac{120}{94.41} = 1.27 \ A \\ \text{Potential difference across, parallel combination} \\ V*=120-(1.27\times 50) = 56.5 \ V \\ Thus, P=\frac{56.5^2}{397}=8.04 \ W$

(c)

x=0.25 R=25

The effective series is we have R1=75 in series with parallel combination of R2=25 and Rbulb=397

$\\ R_{eq}=75+\frac{25\times 397}{397+25} = 98.52 \ \Omega \\ \text{ effective current in the circuit} \\ I=\frac{120}{98.52} = 1.22 \ A \\ \text{Potential difference across, parallel combination} \\ V*=120-(1.22\times 75) = 28.5 \ V \\ Thus, P=\frac{28.5^2}{397}=2.05 \ W$