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

 The given problem involves using a filament bulb designed for a \(100 \, V\) supply in a higher voltage supply of \(230 \, V\). To solve this, we need to find the necessary resistance \(R\) that should be connected in series to make the bulb function properly without damaging it.

1. The power consumed by the bulb is given as \(500 \, W\) when it operates under its designed voltage of \(100 \, V\). Using the power formula, we can find the resistance of the bulb: 

\[P = \frac{V^2}{R_b} \]\]

1. Substituting the given values: 

\[500 = \frac{100^2}{R_b} \] \] Solve f\]

1. The total resistance needed in the circuit for the bulb to function correctly from a \(230 \, V\) supply is found using Ohm’s Law. We know the power and the voltage: 

\[V_{\text{total}} = 230 \, V \] \] Since the bulb consum\]

1. Now, calculate the total resistance required for the circuit with the new supply voltage: 

\[V_{\text{total}} = I \cdot R_{\text{total}} \implies 230 = 5 \cdot R_{\text{total}} \implies R_{\text{total}} = 46 \, \Omega\]

1. Determine the value of the series resistance \(R\) needed: 

\[R_{\text{total}} = R_b + R \implies R = R_{\text{total}} - R_b = 46 - 20 = 26 \, \Omega \] \] Therefore, the required series resistan\]

Thus, the correct answer is \(26 \, \Omega\).