Question

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

• A. 25
• B. 50
• C. 66.6
• D. 100

Answer 1

The Correct Option is B

The total power across the bulb is given as 100 W and the voltage across the bulb is 100 V. The power in a resistor is given by: \[ P = \frac{V^2}{R}. \] For the bulb, we can substitute the known values: \[ 100 = \frac{100^2}{R_{\text{bulb}}}, \] \[ R_{\text{bulb}} = \frac{100^2}{100} = 100 \, \Omega. \] Now, since the bulb is connected across the 400 \( \Omega \) resistor, we calculate the total resistance in the circuit. The equivalent resistance of the series combination is: \[ R_{\text{total}} = 200 + 400 + 100 = 700 \, \Omega. \] Using Ohm's law \( V = IR \), the total current in the circuit is: \[ I = \frac{V_{\text{battery}}}{R_{\text{total}}} = \frac{100}{700} = \frac{1}{7} \, \text{A}. \] The voltage drop across the 400 \( \Omega \) resistor is: \[ V = I \times 400 = \frac{1}{7} \times 400 = 50 \, \text{V}. \]
Final Answer: 50 V