What happens to the power dissipated in a resistor when the emf of the battery connected to it is doubled, while the resistance remains constant?

• A. P/4
• B. P/2
• C. P
• D. 4P

Answer: (D)

When examining how power dissipation in a resistor changes with respect to the electromotive force (emf) provided by a battery, it is essential to understand the relationship given by the formula for electrical power. Power (P) dissipated in a resistor can be expressed through different formulations, but commonly it is formulated as:

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

where ( V ) is the voltage (or emf) across the resistor and ( R ) is the resistance.

If the emf of the battery is doubled while keeping the resistance constant, we can denote the original emf as ( V ) and the new emf as ( 2V ). Plugging this new value into the power formula gives:

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

This simplifies to:

[ P' = \frac{4V^2}{R} ]

Comparing this with the original power ( P = \frac{V^2}{R} ), we see that:

[ P' = 4P ]

This indicates that doubling the emf results in four times the power dissipation in the resistor when resistance remains constant. Thus, the correct answer highlights this