Question

An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

• A. 20 \( \Omega \)
• B. 10 \( \Omega \)
• C. 5 \( \Omega \)
• D. 30 \( \Omega \)

Hint:

When an appliance is used with a series resistor, first calculate the required current and resistance of the appliance from its power and voltage ratings. The series resistor's role is to drop the excess voltage from the supply so that the appliance receives its rated voltage.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a bulb with a specific power and voltage rating. To use it with a higher voltage supply without damaging it, a resistor `R` must be connected in series. We need to find the value of this series resistor.
Step 2: Key Formula or Approach:
1. From the bulb's rating, calculate its resistance (\(R_{bulb}\)) and the current (\(I_{rated}\)) it needs to operate correctly. We can use \(P = V^2/R\) and \(P = VI\).
2. When the bulb and resistor `R` are connected in series to the 200 V supply, the total resistance is \(R_{total} = R_{bulb} + R\).
3. The current flowing through the series circuit must be the bulb's rated current, \(I_{rated}\).
4. Use Ohm's Law for the entire circuit: \(V_{supply} = I_{rated} \times R_{total}\).
Step 3: Detailed Explanation:
Bulb rating: Power \(P_{bulb} = 500\) W, Voltage \(V_{bulb} = 100\) V.
Supply voltage: \(V_{supply} = 200\) V.
Part 1: Find the bulb's properties.
Calculate the resistance of the bulb:
\[ P_{bulb} = \frac{V_{bulb}^2}{R_{bulb}} \implies R_{bulb} = \frac{V_{bulb}^2}{P_{bulb}} = \frac{(100)^2}{500} = \frac{10000}{500} = 20 \, \Omega \] Calculate the current required for the bulb to operate at 500 W:
\[ P_{bulb} = V_{bulb} \times I_{rated} \implies I_{rated} = \frac{P_{bulb}}{V_{bulb}} = \frac{500}{100} = 5 \, \text{A} \] Part 2: Analyze the series circuit.
For the bulb to operate at its rated power, the current flowing through it must be 5 A. Since the resistor R is in series, the same current flows through it.
The total resistance of the circuit is \(R_{total} = R_{bulb} + R = 20 + R\).
According to Ohm's Law for the whole circuit:
\[ V_{supply} = I_{rated} \times R_{total} \] \[ 200 = 5 \times (20 + R) \] Divide by 5:
\[ 40 = 20 + R \] Solve for R:
\[ R = 40 - 20 = 20 \, \Omega \] Step 4: Final Answer:
The resistance to be connected in series is 20 \( \Omega \). This corresponds to option (A).