Question

An electric lamp marked 100 volt d.c. consumes a current of 10A. If it is connected to a 200 volt, 50 Hz AC mains. Calculate the inductance of choke (inductor coil) required :-

• A. 5.5 × $10^{-2}$ H
• B. 5.5 × $10^{-1}$ H
• C. 11.0 H
• D. 11.0 × $10^{2}$ H

Answer 1

Answer: (A)

5.5 × $10^{-2}$ H

Answer 2

1. Determine Lamp Resistance: The lamp's resistance is calculated using its DC rating: $R = \frac{V_{DC}}{I_{DC}} = \frac{100 \, \text{V}}{10 \, \text{A}} = 10 \, \Omega$.
2. Assume AC Current: Assume the RMS current drawn by the lamp in the AC circuit is the same as its DC rating, $I_{AC} = 10 \, \text{A}$.
3. Voltage Drop across Lamp: The voltage drop across the lamp (acting as a resistor) in the AC circuit is $V_R = I_{AC} \times R = 10 \, \text{A} \times 10 \, \Omega = 100 \, \text{V}$.
4. Voltage Drop across Choke: The AC mains voltage is $V_{AC} = 200 \, \text{V}$. Since the lamp and choke are in series, the total voltage is the phasor sum of the voltage across the lamp ($V_R$) and the voltage across the choke ($V_L$): $V_{AC}^2 = V_R^2 + V_L^2$. Substituting the values: $(200 \, \text{V})^2 = (100 \, \text{V})^2 + V_L^2$. $40000 = 10000 + V_L^2$ $V_L^2 = 30000$ $V_L = \sqrt{30000} = 100\sqrt{3} \, \text{V}$.
5. Inductive Reactance: The voltage drop across the choke is $V_L = I_{AC} \times X_L$. $100\sqrt{3} \, \text{V} = 10 \, \text{A} \times X_L$ $X_L = \frac{100\sqrt{3}}{10} \, \Omega = 10\sqrt{3} \, \Omega$.
6. Inductance Calculation: The inductive reactance is $X_L = 2\pi fL$. With $f = 50 \, \text{Hz}$: $10\sqrt{3} = 2\pi \times 50 \times L$ $10\sqrt{3} = 100\pi L$ $L = \frac{10\sqrt{3}}{100\pi} = \frac{\sqrt{3}}{10\pi} \, \text{H}$. Using $\sqrt{3} \approx 1.732$ and $\pi \approx 3.14$: $L \approx \frac{1.732}{10 \times 3.14} = \frac{1.732}{31.4} \approx 0.05515 \, \text{H}$, which is approximately $5.5 \times 10^{-2} \, \text{H}$.