Question

A resistor \( R \) dissipates power \( P \) when connected to a generator. If another resistor \( Q \) is put in series with \( R \), the power dissipated by \( R \) will:

• A. Increase
• B. Decrease
• C. Remain the same
• D. Increase or decrease depending on the values of \( R \) and \( Q \)

Hint:

When resistors are connected in series, the total resistance increases, which reduces the current and decreases the power dissipated by individual resistors in the circuit.

Answer 1

The Correct Option is B

Step 1: Power dissipation in a resistor.
The power dissipated by a resistor \( R \) when connected to a voltage \( V \) is given by the formula:
\[ P = \frac{V^2}{R} \]
This formula indicates that the power dissipated depends on the voltage across the resistor and its resistance.
Step 2: Power dissipation when resistors are in series.
When two resistors, \( R \) and \( Q \), are connected in series, the total resistance of the circuit becomes:
\[ R_{{total}} = R + Q \]
The voltage across the entire series combination remains the same as when only \( R \) was connected, but the current in the circuit changes because the total resistance is now higher.
The current \( I \) in the circuit is given by Ohm's law:
\[ I = \frac{V}{R_{{total}}} = \frac{V}{R + Q} \]
Thus, the current decreases because the total resistance is greater.
The power dissipated by \( R \) is now:
\[ P_R = I^2 R = \left( \frac{V}{R + Q} \right)^2 R \]
This shows that the power dissipated by \( R \) will decrease because the current in the circuit is reduced due to the increase in total resistance.
Step 3: Conclusion.
The power dissipated by \( R \) will decrease when resistor \( Q \) is added in series with \( R \), since the total resistance of the circuit increases, reducing the current and hence the power dissipated by \( R \).
Thus, the correct answer is (2) Decrease.