Question

In the circuit shown in the figure, if both the bulbs B₁ and B₂ are identical :-

• A. Their brightness will be the same
• B. B₂ will be brighter than B₁
• C. as frequency of supply voltage is increased, brightness of B₁ will increase & that of B₂ will decrease
• D. only B₂ will glow because the capacitor has infinite impedance

Answer 1

Answer: (B)

B₂ will be brighter than B₁

Answer 2

The circuit consists of an AC voltage source connected to two parallel branches. One branch contains a bulb B₁ in series with a capacitor C. The other branch contains a bulb B₂ in series with an inductor L. Both bulbs are identical, meaning they have the same resistance R. The brightness of a bulb is proportional to the power dissipated by it, which is given by $P = I_{rms}^2 R$. Since R is the same for both bulbs, the brightness is proportional to the square of the RMS current flowing through the bulb.

The RMS voltage of the source is $V_{rms} = 220 V$ and the frequency is $f = 50 Hz$. The angular frequency is $\omega = 2\pi f = 2\pi (50) = 100\pi \, rad/s$. The capacitance is $C = 500 \mu F = 500 \times 10^{-6} F = 5 \times 10^{-4} F$. The inductance is $L = 10 mH = 10 \times 10^{-3} H = 10^{-2} H$.

The impedance of the branch containing bulb B₁ and capacitor C is $Z_1 = R + jX_C$, where $X_C = \frac{1}{\omega C}$ is the capacitive reactance. The magnitude of the impedance is $|Z_1| = \sqrt{R^2 + X_C^2}$. The RMS current through bulb B₁ is $I_{1,rms} = \frac{V_{rms}}{|Z_1|} = \frac{V_{rms}}{\sqrt{R^2 + X_C^2}}$.

The impedance of the branch containing bulb B₂ and inductor L is $Z_2 = R + jX_L$, where $X_L = \omega L$ is the inductive reactance. The magnitude of the impedance is $|Z_2| = \sqrt{R^2 + X_L^2}$. The RMS current through bulb B₂ is $I_{2,rms} = \frac{V_{rms}}{|Z_2|} = \frac{V_{rms}}{\sqrt{R^2 + X_L^2}}$.

Let's calculate the reactances at $f = 50 Hz$: $X_C = \frac{1}{\omega C} = \frac{1}{(100\pi)(5 \times 10^{-4})} = \frac{1}{5\pi \times 10^{-2}} = \frac{100}{5\pi} = \frac{20}{\pi} \approx 6.366 \Omega$. $X_L = \omega L = (100\pi)(10^{-2}) = \pi \Omega \approx 3.142 \Omega$.

Comparing the reactances, we have $X_C > X_L$. Since $X_C > X_L$, we have $R^2 + X_C^2 > R^2 + X_L^2$, which means $|Z_1| > |Z_2|$. Since the voltage across both parallel branches is the same ($V_{rms}$), the current through the branch with lower impedance will be higher. Thus, $I_{1,rms} = \frac{V_{rms}}{|Z_1|} < \frac{V_{rms}}{|Z_2|} = I_{2,rms}$. Since $I_{1,rms} < I_{2,rms}$, the power dissipated by B₁ will be less than the power dissipated by B₂, and hence B₁ will be less bright than B₂. So, B₂ will be brighter than B₁.