Question

Two resistors of \(200\,\Omega\) and \(400\,\Omega\) are connected in series with a battery of \(100\,\text{V}\). A bulb rated at \(200\,\text{V}, 100\,\text{W}\) is connected across the \(400\,\Omega\) resistance. The potential drop across the bulb is _________ V.

• A. \(25\)
• B. \(50\)
• C. \(66.6\)
• D. \(100\)

Hint:

Remember:

• Bulb resistance: \[ R = \frac{V^2}{P} \]
• Same voltage across parallel combination
• In parallel: \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \]

Answer 1

The Correct Option is B

Concept: When a bulb is connected across a resistor, both become parallel combinations. First, the resistance of the bulb is calculated using: \[ R = \frac{V^2}{P} \] Then equivalent resistance is found using parallel combination formula.

Step 1: Calculate the resistance of the bulb. Given: \[ V = 200\,\text{V}, \qquad P = 100\,\text{W} \] Using: \[ R = \frac{V^2}{P} \] \[ R_b = \frac{(200)^2}{100} \] \[ R_b = \frac{40000}{100} \] \[ R_b = 400\,\Omega \]

Step 2: Find equivalent resistance of bulb and \(400\,\Omega\) resistor. The bulb is connected across the \(400\,\Omega\) resistor. Hence: \[ 400\,\Omega \parallel 400\,\Omega \] Using parallel combination: \[ R_p = \frac{400 \times 400}{400 + 400} \] \[ R_p = \frac{160000}{800} \] \[ R_p = 200\,\Omega \]

Step 3: Find total resistance of the circuit. Now this equivalent resistance is in series with the \(200\,\Omega\) resistor. \[ R_{\text{total}} = 200 + 200 \] \[ R_{\text{total}} = 400\,\Omega \]

Step 4: Calculate the circuit current. Using Ohm’s law: \[ I = \frac{V}{R} \] \[ I = \frac{100}{400} \] \[ I = 0.25\,\text{A} \]

Step 5: Find the potential drop across the parallel combination. Voltage across parallel branch: \[ V_p = I \times R_p \] \[ V_p = 0.25 \times 200 \] \[ V_p = 50\,\text{V} \] Since the bulb is connected across this branch, the potential drop across the bulb is: \[ \boxed{50\,\text{V}} \]