Question

The resistance of the tungsten wire in the light bulb, which is rated at $120$ V/$75$ W and powered by a $120$ V direct-current supply, is

• A. $0.37\,\Omega$
• B. $1.2\,\Omega$
• C. $2.66\,\Omega$
• D. $192\,\Omega$
• E. $9 \times 10^3\,\Omega$

Hint:

For electrical appliances, use $R = \frac{V^2}{P}$ when voltage and power rating are given.

Answer 1

The Correct Option is D

Concept:
Power relation: \[ P = \frac{V^2}{R} \]

Step 1: Rearrange formula.
\[ R = \frac{V^2}{P} \]

Step 2: Substitute values.
\[ R = \frac{120^2}{75} \]

Step 3: Calculate.
\[ R = \frac{14400}{75} = 192\,\Omega \]