Question

A bulb $B$ is connected to a source having constant emf and some internal resistance. A variable resistance $R$ is connected in parallel to the bulb. If the resistance $R$ is increased, how will it affect the

(a) Brightness of the bulb? (b) Power spent by the source?

• A. (a) Brightness of the bulb: Increases, (b) Power spent by the source: Decreases
• B. (a) Brightness of the bulb: Decreases, (b) Power spent by the source: Increases
• C. (a) Brightness of the bulb: Increases, (b) Power spent by the source: Increases
• D. (a) Brightness of the bulb: Decreases, (b) Power spent by the source: Decreases

Answer 1

Answer: (A)

(a) Brightness of the bulb: Increases, (b) Power spent by the source: Decreases

Answer 2

(a) Brightness of the bulb:

1. The equivalent resistance of the parallel combination of the bulb ($R_B$) and the variable resistor ($R$) is $R_{parallel} = \frac{R_B R}{R_B + R}$. As $R$ increases, $R_{parallel}$ increases.
2. The voltage across the parallel combination (and hence across the bulb) is $V_{parallel} = E \frac{R_{parallel}}{r + R_{parallel}}$, where $E$ is the constant EMF and $r$ is the internal resistance. As $R_{parallel}$ increases, $V_{parallel}$ increases.
3. The power dissipated by the bulb, which determines its brightness, is $P_B = \frac{V_{parallel}^2}{R_B}$. Since $V_{parallel}$ increases and $R_B$ is constant, $P_B$ increases. Thus, the brightness of the bulb increases.

(b) Power spent by the source:

1. The total current drawn from the source is $I = \frac{E}{r + R_{parallel}}$. As $R_{parallel}$ increases, the denominator $(r + R_{parallel})$ increases, causing the total current $I$ to decrease.
2. The power spent by the source is $P_{source} = E \times I$. Since $E$ is constant and $I$ decreases, the power spent by the source decreases.