Question

A lamp is rated at 240V, 60W. When in use the resistance of the filament of the lamp is 20 times that of the cold filament. The resistance of the lamp when not in use is:

• A. \( 54 \, \Omega \)
• B. \( 60 \, \Omega \)
• C. \( 50 \, \Omega \)
• D. \( 48 \, \Omega \)

Hint:

To find the resistance of a filament when it is not in use, use the relationship between power and resistance, and apply the given ratio of resistances in the two cases.

Answer 1

The Correct Option is D

To determine the resistance of the lamp when not in use, we follow these steps:

1. Given the power rating (P) and voltage rating (V), we can find the resistance of the lamp when in use using the formula:
\( R = \frac{V^2}{P} \)

2. Substituting the given values \( V = 240 \, \text{V} \) and \( P = 60 \, \text{W} \):
\( R = \frac{240^2}{60} = \frac{57600}{60} = 960 \, \Omega \)

3. It's given that when the lamp is in use, the resistance of the filament is 20 times the resistance of the cold filament:

\( R_{\text{hot}} = 20 \times R_{\text{cold}} \)

4. We know from step 2 that \( R_{\text{hot}} = 960 \, \Omega \). Solving for \( R_{\text{cold}} \):
\( 960 \, \Omega = 20 \times R_{\text{cold}} \)
\( R_{\text{cold}} = \frac{960 \, \Omega}{20} = 48 \, \Omega \)

Thus, the resistance of the lamp when not in use is 48 \( \Omega \).

Answer 2

We are given the lamp's rated power \( P = 60 \, {W} \) and the voltage across it \( V = 240 \, {V} \). The resistance of the filament when the lamp is in use is 20 times the resistance when it is not in use. We need to find the resistance of the lamp when it is not in use. 
Step 1: Using the formula for power, \( P = \frac{V^2}{R_{{in use}}} \), where \( R_{{in use}} \) is the resistance of the lamp when it is in use, we can solve for \( R_{{in use}} \): \[ R_{{in use}} = \frac{V^2}{P} = \frac{240^2}{60} = 960 \, \Omega. \] 
Step 2: We are told that the resistance when the lamp is in use is 20 times the resistance when it is not in use, i.e., \[ R_{{in use}} = 20 \times R_{{not in use}}. \] Substitute the value of \( R_{{in use}} \): \[ 960 = 20 \times R_{{not in use}}. \] Solving for \( R_{{not in use}} \): \[ R_{{not in use}} = \frac{960}{20} = 48 \, \Omega. \] Thus, the resistance of the lamp when not in use is \( 48 \, \Omega \).