Question

A lamp “50 watt and 100 volt” is to be connected to AC mains of 200 volt 50 Hz. Calculate the capacity of condenser required in series of lamp.

Hint:

In series AC circuits, voltages across different components do not add up arithmetically. They add as vectors (phasors). The relation is \( V_S^2 = V_R^2 + V_C^2 \). You could also solve this problem by finding \(V_C = \sqrt{V_S^2 - V_R^2}\), then \(X_C = V_C/I\), and finally C from \(X_C\).

Answer 1

Step 1: Understanding the Problem and Calculating Lamp Parameters:
The lamp is designed to work at 100 V, but the AC source is 200 V. To operate the lamp safely, we must connect a component in series to drop the excess voltage. Here, a capacitor (condenser) is used. The lamp itself behaves as a resistor.
First, let's find the current required by the lamp and its resistance from its ratings.
Power of the lamp, \( P = 50 \) W
Voltage rating of the lamp, \( V_R = 100 \) V
The current (RMS) that must flow through the lamp for it to work properly is: \[ I = \frac{P}{V_R} = \frac{50 \, \text{W}}{100 \, \text{V}} = 0.5 \, \text{A} \] This will be the current flowing through the entire series RC circuit.
The resistance of the lamp is: \[ R = \frac{V_R}{I} = \frac{100 \, \text{V}}{0.5 \, \text{A}} = 200 \, \Omega \] Step 2: Analyzing the Series RC Circuit:
We now have a circuit with a resistor (the lamp, R=200 \(\Omega\)) and a capacitor (C) connected in series to an AC source.
Source voltage, \( V_S = 200 \) V
Circuit current, \( I = 0.5 \) A
Frequency, \( f = 50 \) Hz
The total impedance (Z) of the circuit can be calculated as: \[ Z = \frac{V_S}{I} = \frac{200 \, \text{V}}{0.5 \, \text{A}} = 400 \, \Omega \] Step 3: Calculating Capacitive Reactance (\(X_C\)):
The impedance of a series RC circuit is given by the formula: \[ Z = \sqrt{R^2 + X_C^2} \] where \(X_C\) is the capacitive reactance.
We can solve for \(X_C\): \[ Z^2 = R^2 + X_C^2 \] \[ (400)^2 = (200)^2 + X_C^2 \] \[ 160000 = 40000 + X_C^2 \] \[ X_C^2 = 120000 \] \[ X_C = \sqrt{120000} = \sqrt{40000 \times 3} = 200\sqrt{3} \, \Omega \approx 346.4 \, \Omega \] Step 4: Calculating Capacitance (C):
The capacitive reactance is related to the capacitance and frequency by the formula: \[ X_C = \frac{1}{2\pi f C} \] Rearranging to solve for C: \[ C = \frac{1}{2\pi f X_C} \] Substituting the values: \[ C = \frac{1}{2\pi (50)(200\sqrt{3})} = \frac{1}{20000\pi\sqrt{3}} \] \[ C \approx \frac{1}{20000 \times 3.14 \times 1.732} \approx \frac{1}{108828} \approx 9.2 \times 10^{-6} \, \text{F} \] This is equal to 9.2 microfarads (\(\mu\)F).
Step 5: Final Answer:
The required capacity of the condenser is \( 9.2 \times 10^{-6} \) F or 9.2 \(\mu\)F.