Question

The resistor $R_1 = 3 \ \Omega$ and $R_2 = 1 \ \Omega$ are connected in parallel to a 20 V battery. Find the heat developed in the resistor $R_1$ in one minute

• A. 600 J
• B. 800 J
• C. 6000 J
• D. 8000 J
• E. 7000 J

Hint:

When resistors are in parallel, use $H = \frac{V^2}{R}t$ because the voltage is the same. When they are in series, use $H = I^2Rt$ because the current is the same.

Answer 1

The Correct Option is D

Concept: For resistors in parallel, the voltage across each resistor is equal to the source voltage. The heat produced ($H$) in a resistor is given by Joule's law of heating: $$H = \frac{V^2}{R}t$$ where $V$ is voltage, $R$ is resistance, and $t$ is time in seconds.

Step 1: {Identify the voltage across $R_1$.}
Since $R_1$ and $R_2$ are in parallel with the 20 V battery, the voltage across $R_1$ is: $$V_1 = 20 \text{ V}$$

Step 2: {Convert time to S.I. units.}
The time given is one minute. $$t = 1 \times 60 = 60 \text{ seconds}$$

Step 3: {Calculate the heat developed in $R_1$.}
Using the formula $H = \frac{V^2}{R_1}t$: $$H = \frac{(20)^2}{3} \times 60$$ $$H = \frac{400}{3} \times 60$$ $$H = 400 \times 20 = 8000 \text{ J}$$