A 6V battery is connected to a 3 ohm resistor. What is the energy dissipated in 2 seconds?
Q1. A 6V battery is connected to a 3 ohm resistor. What is the energy dissipated in 2 seconds?
Answer: 24 J
Explanation: First, find the current I = V/R = 6/3 = 2 A. Then, using the formula E = I^2Rt, we get E = 2^2 * 3 * 2 = 24 J. Option A is incorrect because it underestimates the energy.