Question

A circuit is constructed with five resistors and one real battery as shown above right. We model. The real battery as an ideal emf V = 12 V in series with an internal resistance r as shown above left. The values for the resistors are: R = R3 = 40 N, R4 = Rs = 80 and R2 = 142 N. The measured voltage across the terminals of the batery is Vbattery = 11.41 V. I R R R R4 R4 R R V R5 Ra + ün 1) What is 11, the current that flows through the resistor R? mA Submit + 2) What is r, the internal resistance of the battery? 2 Submit + 3) What is 13, the current through resistor Rz? mA Submit 4) What is P2, the power dissipated in resistor Ry? W Submit + 5) What is V2, the magnitude of the voltage across the resistor R2? v Submit + 6) Resistor R2 is now shorted out as shown. How does the magnitude of the voltage across the battery change? I, R R RA R R5 Vbattery decreases Vbattery increases Vbattery remains the same Submit

Answer 1

valent. Ri= R3 = 40, Rey=85= 80, R2 = 142 V Bart = 11.4iv Vemf=12 =40+80+80=2002 y Rz, Ry & Rg in series 100+ 142 in series =123.04 +8 23.04ot notr on by 12 * 200 r is in parallel with R2 u P=12 R2 =(0.092-0.038) :(142) = 0.417W Answer For Requi 2 08 142 = 83.040 r Rox 83.000, R. &r Rey - Current drawin i- Rey 123.outo Battery. of Battery 12 1) I,=i= 0.09273A = 92.73 m A 123.84 +6.36 Answes 2) 8=6.36 cars was R2 142x0.092 I, =4477 0.03850 Qq + R3t Ruths 142+40+80+80 = 38.50mA Arom Now Voltage drop across &=12-11:41=0.5gr. .8 Y=cir *0.59= 12. (123:04+0) 8=6362 Internal Resistance 3) 13" lla 12 not 6.36 Vg = IR = 0.05 X 142 = 7.70101¢ Answer = V=IR = 0.258 (ho) = 10.35 v 10.35<11.04) > Voltage of Battery Decreases. = 0258 1