The current in the figure below is 0.69 A. (Let the internal resistance of batte

Question

The current in the figure below is 0.69 A. (Let the internal resistance of battery 1 be
r1 = 1.2 and the internal resistance of battery 2 be
r2 = 1.1 .)

The current in the figure below is 0.69 A. (Let the internal resistance of battery 1 be r1 = 1.2 ohm and the internal resistance of battery 2 be r2 = 1.1 ohm.) (a) Find the voltage drop Vab. (b) Find the power dissipated in R. (c) Find the useful electric power supplied by battery 2. (Note: Do not forget the internal resistance of battery 2.) (d) Find the rate at which chemical energy is stored in battery 1. (Note: Do not include power dissipated in the internal resistor of battery 1 in your answer.)

Explanation / Answer

3/R+2.3 = 0.69 ==> R = 2.04 ohms hence Vab = R * 0.69 = 2.04 * 0.69 = 1.413 volts power in R = i^2 R = 0.974 watts useful energy by battery 2 = Vi - r2i^2 = = 9 * 0.69 - 1.1 * 0.69 *0.69 = 5.68 watts rate of energy stortage = Vi = 6 * 0.69 = 4.14 watts

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