Question

1. Suppose the voltage output of a battery is 24.0V in a series circuit, and the resistances are Ri=2.0012, R2=6.0092, and R3=10.092. a. Draw a diagram of the circuit. b. What is the total resistance and current at each "spot" in the circuit? C. Calculate the voltage drop in each resistor. d. Calculate the power dissipated by each resistor and the power output of the source. 2. Suppose the voltage output of a battery is 24.0V in a parallel circuit, and the resistances are Ri=2.0012, R2=6.0022, and R3=10.092. a. Draw a diagram of the circuit. b. What is the total resistance and current at each "spot" in the circuit? c. Calculate the voltage drop in each resistor. d. Calculate the power dissipated by each resistor and the power output of the source.

Answer 1

Ri R2 R3 And 1. (a) th 24 Volt 18 (6) total Resistance R = R1 + R2 + R₂ = 2 + 640= 182 current through battery I2 24 = 1.33 Ant (C) Voltage drop at R, a V, = 1:33x2 = 2.66 Vold Voltage drop at R2 ² % 2 13376 - 8.00 Volt Voltage drop at R3 = V₂ = 1:33x10 = 13.33 Volt (d) power output of source PE 32 welt 18 P = I'R = 933²42 = 3:55 watt P2 = 1.332x6 10.66 watt 1,33² x 10 = 17.77 watt 242 R P3 =

RI A 2. (a) R, RS 24 NOH (6) total Resistance s (8 + Best) = 1.32 Current through Ry - Is Y = 24 - 12 Amp Ri 2 I = 24 = 4 Amp 2 24 zu - 2.4 Amp (e) Yoltage drop at R, = N = 24 Volt N2 = 2y volt 1/2 = 24 Volt (d) 24² - 441.7 watt "R 1.3 P = y² LE Pi = 242 S 288 watt P 2 242 6 M 96 watt P3 11 242 = 576 watt