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

Step 1: Understanding the Concept:
The bulb acts as a resistor in the circuit. We first need to find its resistance using its power rating.
Connecting the bulb across the \(400\ \Omega\) resistor places them in parallel.
We then find the equivalent resistance of the entire circuit to find the main current, which will allow us to calculate the voltage drop across the parallel combination.
Step 2: Key Formula or Approach:
Resistance from power rating: \(R = \frac{V_{rated}^2}{P_{rated}}\).
Equivalent resistance for two parallel resistors: \(R_p = \frac{R_1 R_2}{R_1 + R_2}\).
Ohm's Law: \(V = I R_{eq}\).
Step 3: Detailed Explanation:
Calculate the resistance of the bulb:
\[ R_{bulb} = \frac{(200)^2}{100} = \frac{40000}{100} = 400\ \Omega \] The bulb is connected in parallel with the \(400\ \Omega\) resistor. Calculate the equivalent resistance of this parallel pair:
\[ R_p = \frac{400 \times 400}{400 + 400} = \frac{160000}{800} = 200\ \Omega \] This parallel combination is in series with the \(200\ \Omega\) resistor. Calculate the total resistance of the circuit:
\[ R_{total} = 200\ \Omega + R_p = 200 + 200 = 400\ \Omega \] Now, calculate the total current drawn from the 100 V battery:
\[ I_{total} = \frac{V_{battery}}{R_{total}} = \frac{100}{400} = 0.25 \text{ A} \] The potential drop across the bulb is the same as the potential drop across the entire parallel combination.
\[ V_{bulb} = I_{total} \times R_p = 0.25 \times 200 = 50 \text{ V} \] Step 4: Final Answer:
The potential drop across the bulb is \(50 \text{ V}\).