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

Step 1: Write down the given data.
Lamp A: Power \( P_A = 50 \, \text{W} \), Voltage rating \( V_A = 220 \, \text{V} \)
Lamp B: Power \( P_B = 100 \, \text{W} \), Voltage rating \( V_B = 220 \, \text{V} \)
Step 2: Recall the formula relating power, voltage, and resistance.
For an electrical device, the power is given by: \[ P = \frac{V^2}{R} \] where \( V \) is the voltage rating and \( R \) is the resistance of the device.
Step 3: Calculate the resistance of each lamp.
From \( P = \frac{V^2}{R} \), we get: \[ R = \frac{V^2}{P} \] For lamp A: \[ R_A = \frac{V_A^2}{P_A} = \frac{(220)^2}{50} \] For lamp B: \[ R_B = \frac{V_B^2}{P_B} = \frac{(220)^2}{100} \]
Step 4: Find the ratio \( R_A : R_B \).
\[ \frac{R_A}{R_B} = \frac{\frac{(220)^2}{50}}{\frac{(220)^2}{100}} = \frac{(220)^2}{50} \times \frac{100}{(220)^2} \] The \( (220)^2 \) cancels out: \[ \frac{R_A}{R_B} = \frac{100}{50} = \frac{2}{1} \]
Step 5: Simplify the ratio.
\[ R_A : R_B = 2 : 1 \]
Step 6: Interpretation.
Lamp A (50 W) has twice the resistance of lamp B (100 W). This makes sense because for the same voltage rating, a lower power lamp has higher resistance.
Step 7: Final answer.
\[ \boxed{R_A : R_B = 2 : 1} \]