Question

Two lamps 'A' and 'B' of rating 50 W; 220 V and 100 W, 220 V are connected in series combination. Find out the ratio of the resistances (\(R_A : R_B\)) of these lamps.

Hint:

For devices with same voltage rating, resistance is inversely proportional to power: \( R \propto \frac{1}{P} \). So \( \frac{R_A}{R_B} = \frac{P_B}{P_A} = \frac{100}{50} = 2 : 1 \).

Answer 1

Given:
Lamp A: 50 W, 220 V
Lamp B: 100 W, 220 V

Step 1: Use the formula relating Power, Voltage and Resistance
\[ P = \frac{V^2}{R} \] Rearranging to find resistance: \[ R = \frac{V^2}{P} \]

Step 2: Calculate Resistance of Lamp A \[ R_A = \frac{220^2}{50} \] \[ R_A = \frac{48400}{50} \] \[ R_A = 968 \, \Omega \]

Step 3: Calculate Resistance of Lamp B \[ R_B = \frac{220^2}{100} \] \[ R_B = \frac{48400}{100} \] \[ R_B = 484 \, \Omega \]

Step 4: Find the Ratio \[ R_A : R_B = 968 : 484 \] Divide both by 484: \[ R_A : R_B = 2 : 1 \]

Final Answer: \[ \boxed{R_A : R_B = 2 : 1} \]
The 50 W lamp has greater resistance because for the same voltage, lower power corresponds to higher resistance.