Question

A resistance R is connected across an ideal battery. The total power dissipated in the circuit is P. If another resistance R is added in series,the new total dissipated power is:

• A. 2P
• B. 4P
• C. P
• D. \(\frac{P}{2}\)
• E. \(\frac{P}{4}\)

Answer 1

The Correct Option is D

Given:

• Initial setup: A single resistor \( R \) connected to an ideal battery (voltage \( V \)).
• Power dissipated initially: \( P \).
• Modified setup: Another resistor \( R \) added in series with the first one.

Step 1: Calculate Initial Power (\( P \))

For the initial single-resistor circuit:

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

Step 2: Analyze Modified Circuit

When a second resistor \( R \) is added in series:

Total resistance becomes:

\[ R_{\text{total}} = R + R = 2R \]

The new power dissipated (\( P_{\text{new}} \)) is:

\[ P_{\text{new}} = \frac{V^2}{R_{\text{total}}} = \frac{V^2}{2R} \]

Step 3: Compare Powers

Substitute the initial power \( P = \frac{V^2}{R} \):

\[ P_{\text{new}} = \frac{V^2}{2R} = \frac{1}{2} \left( \frac{V^2}{R} \right) = \frac{P}{2} \]

Conclusion:

The new total dissipated power is \( \frac{P}{2} \).

Answer: \(\boxed{D}\)

Answer 2

Step 1: Recall the formula for power dissipated in a circuit.

The power dissipated in a circuit is given by:

\[ P = \frac{V^2}{R_{\text{total}}}, \]

where:

• \( V \) is the voltage of the battery, and
• \( R_{\text{total}} \) is the total resistance in the circuit.

 

In the initial case, there is only one resistor \( R \) connected across the battery. Thus, the total resistance is \( R_{\text{total}} = R \), and the power dissipated is:

\[ P = \frac{V^2}{R}. \]

Step 2: Analyze the effect of adding another resistor in series.

When another resistor \( R \) is added in series, the total resistance becomes:

\[ R_{\text{total}} = R + R = 2R. \]

The new power dissipated in the circuit is:

\[ P_{\text{new}} = \frac{V^2}{R_{\text{total}}} = \frac{V^2}{2R}. \]

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

\[ P_{\text{new}} = \frac{1}{2} \cdot \frac{V^2}{R} = \frac{P}{2}. \]

Final Answer: The new total dissipated power is \( \mathbf{\frac{P}{2}} \), which corresponds to option \( \mathbf{(D)} \).