In the circuit shown below, the voltages of the batteries are V_1 = 52 Volts, V_
ID: 1595897 • Letter: I
Question
In the circuit shown below, the voltages of the batteries are V_1 = 52 Volts, V_2 = 4 Volts, and V_3 = 11 Volts. To setup the problem, battery 1 (V_1) is used to guess the initial direction of current flow. Call I the current out of the battery. I_1 the current traveling through R_2, and I_2 the current traveling through R_5 (as was done in class). When the correct loop and function equations are solved for the currents, I_1 = 15 amps and I_2 = -5 amps. What is the total power dissipated by all the resistors in Watts?Explanation / Answer
I = I2 + I1
I = 15 - 5 = 10 amps
power dissipated = power across v1 , v2 , v3
P = P1 + P2 + P3
P = V1*I - V2*I1 - V3*I2
P = (52*10) - (4*15) - (11*5)
P = 405 W