Question

A filament bulb $(500\, W, 100 \,V)$ is to be used in $230\, V$ main supply. When a resistance $R$ is connected in series, it works perfectly and the bulb consumes $500\, W$. The value of $R$ is

• A. $230\, \Omega $
• B. $46\, \Omega $
• C. $26 \, \Omega $
• D. $13 \, \Omega $

Answer 1

The Correct Option is C

Power, Resistance, and Current Calculation 

We are given the following formula for power:

\(P = \frac{V^2}{R}\)

Step 1: Calculating the Resistance (\( R_b \))

Using the formula for power, we can calculate the resistance \( R_b \) as:

\(R_b = \frac{V^2}{P} = \frac{10000}{500} = 20 \, \Omega\)

Step 2: Using Ohm's Law to Find Current

Next, we use Ohm's law to find the current. From Ohm's law, we know:

\(i = \frac{V}{R_b} = \frac{130}{R}\)

Step 3: Equating the Currents

We now set up the equation by equating the current expressions:

\(\frac{100}{20} = \frac{130}{R}\)

Step 4: Solving for \( R \)

Solving for \( R \), we get:

\(R = \frac{130 \times 20}{100} = 26 \, \Omega\)

Conclusion:

The resistance \( R \) is \( 26 \, \Omega \).