Question

The resistors $R_1 = 3\Omega$ 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:

For parallel circuits, use $H = \frac{V^2}{R}t$ because $V$ is constant. For series circuits, use $H = I^2Rt$ because $I$ is constant.

Answer 1

The Correct Option is D

To find the heat developed in resistor \( R_1 \) over one minute, we begin by calculating the current through \( R_1 \) given that \( R_1 \) and \( R_2 \) are connected in parallel to a 20 V battery.

Find the total current from the battery:

The voltage across each resistor in a parallel circuit is the same as the source voltage. Thus, the voltage across \( R_1 \) and \( R_2 \) is 20 V each.

Calculate the current through each resistor using Ohm's Law:

The current through \( R_1 \) is given by Ohm's Law: \(I_1 = \frac{V}{R_1}\).

Substitute the given values: \( I_1 = \frac{20 \, \text{V}}{3 \, \Omega} \). Thus, \(I_1 = \frac{20}{3} \approx 6.67 \, \text{A}\).

Calculate the heat developed in \( R_1 \):

The heat developed \( H \) in a resistor over time is given by the formula: \(H = I^2 R t\), where \( I \) is the current through the resistor, \( R \) is the resistance, and \( t \) is the time in seconds.

Given that \( t = 60 \) seconds (1 minute), substitute the known values: \( H = (6.67 \, \text{A})^2 \times 3 \, \Omega \times 60 \, \text{s} \).

Calculate: \(H = 6.67^2 \times 3 \times 60 \approx 8000 \, \text{J}\).

Conclusion:

The heat developed in the resistor \( R_1 \) in one minute is \(8000 \, \text{J}\).