Question

Across the \( 220 \, V \) source of internal resistance \( 20\Omega \), how many lamps of \( 40 \, W, 100 \, V \) can be connected in parallel so that all the lamps may glow with full brightness.

• A. 40
• B. 15
• C. 30
• D. 20

Hint:

Always ensure rated voltage across each appliance for full brightness, then use internal resistance to find total current.

Answer 1

The Correct Option is B

Step 1: Find resistance of one lamp.
Using \( P = \frac{V^2}{R} \):
\[ R = \frac{V^2}{P} = \frac{100^2}{40} = \frac{10000}{40} = 250\Omega \]

Step 2: Condition for full brightness.
Each lamp must receive \( 100 \, V \).

Step 3: Voltage drop in internal resistance.
Total voltage = \( 220V \), so drop across internal resistance:
\[ 220 - 100 = 120V \]

Step 4: Find total current drawn.
\[ I = \frac{120}{20} = 6A \]

Step 5: Current drawn by one lamp.
\[ I_1 = \frac{P}{V} = \frac{40}{100} = 0.4A \]

Step 6: Find number of lamps.
\[ n = \frac{I}{I_1} = \frac{6}{0.4} = 15 \]

Step 7: Final Answer.
\[ \boxed{15} \]