Question

An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is :

• A. 12.5 W
• B. 25 W
• C. 50 W
• D. 100 W

Answer 1

The Correct Option is A

Step 1: Finding Resistance (\(R\))

We use the formula for resistance when the rated voltage (\(V\)) and rated power (\(P\)) are given:
\(R = \frac{V^2}{P}\)

Given:
\(V = 200 \, \text{volts}, \quad P = 50 \, \text{watts}\)

Substituting the values:
\(R = \frac{200^2}{50} = \frac{40000}{50} = 800 \, \Omega\)

Thus, the resistance is:
\(R = 800 \, \Omega\)

Step 2: Finding Power (\(P\)) for a Different Voltage

To calculate the power consumed for an applied voltage (\(V_{\text{applied}}\)) of 100 volts, we use the formula:
\(P = \frac{V_{\text{applied}}^2}{R}\)

Given:
\(V_{\text{applied}} = 100 \, \text{volts}, \quad R = 800 \, \Omega\)

Substituting the values:
\(P = \frac{100^2}{800} = \frac{10000}{800} = 12.5 \, \text{watts}\)

Thus, the power consumed is:
\(P = 12.5 \, \text{watts}\)

Answer 2

Step 1: Understand what the rating means
A bulb rated 50 W – 200 V means: when the potential difference across the bulb is 200 V, it draws just enough current so that the electrical power converted to heat and light is 50 W. This rating allows us to determine the bulb’s intrinsic resistance, which we then treat as constant (ideal filament approximation at fixed temperature) to predict behavior at other voltages.

Step 2: Extract the bulb’s resistance from its rating
Use the power–voltage–resistance relation \(P=\dfrac{V^2}{R}\). At the rated condition:
\[ P_{\text{rated}}=\frac{V_{\text{rated}}^2}{R}\quad\Longrightarrow\quad R=\frac{V_{\text{rated}}^2}{P_{\text{rated}}} \] Substitute \(V_{\text{rated}}=200\ \text{V}\) and \(P_{\text{rated}}=50\ \text{W}\):
\[ R=\frac{(200)^2}{50}=\frac{40000}{50}=800\ \Omega \] So the bulb’s resistance is \(R=800\ \Omega\).

Step 3: Compute power at the new supply voltage
Now the bulb is connected across \(V_{\text{new}}=100\ \text{V}\). Assuming the same resistance \(R\approx800\ \Omega\), the power dissipated is
\[ P_{\text{new}}=\frac{V_{\text{new}}^2}{R}=\frac{(100)^2}{800}=\frac{10000}{800}=12.5\ \text{W} \] Thus, the bulb dissipates \(12.5\ \text{W}\) at 100 V.

Step 4: Quick proportionality check (sanity check)
For a fixed \(R\), power scales with the square of voltage: \(P\propto V^2\). Halving the voltage from 200 V to 100 V reduces power by a factor of \((\tfrac{1}{2})^2=\tfrac{1}{4}\). A quarter of 50 W is \(50\times\tfrac{1}{4}=12.5\ \text{W}\). This matches our calculation.

Step 5: Unit and concept checks
Units: Using \(P=V^2/R\) gives \(\text{V}^2/\Omega = \text{W}\), which is consistent.
Model note: Real incandescent filaments change resistance with temperature. At lower voltage, the filament is cooler and the true resistance would be slightly smaller, leading to slightly different power. However, textbook problems assume constant \(R\) from the rated condition unless stated otherwise; the proportional \(V^2\) law applies here.

Common pitfalls to avoid
1) Using \(P=VI\) without first finding the correct current at the new voltage; or mixing rated-current with new voltage.
2) Assuming power scales linearly with voltage; it actually scales as \(V^2\) for fixed \(R\).

Final answer

12.5 W