What is r, the internal resistance of the battery? What is I 3 , the current thr

Question

What is r, the internal resistance of the battery?

What is I3, the current through resistor R3?

What is P2, the power dissipated in resistor R2?

What is V2, the magnitude of the voltage across the resistor R2?

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: R1-R3 38 , R4-R5 = 77 and R2-1370. The measured voltage across the terminals of the batery is Vbattery = 11.42 V. I, R I R I R I2 R R4 R4 R2 R2 Rs R5

Explanation / Answer

R3 R4 R5 in series so that

R3+R4+R5 = 38+77+77 =192 =R345

R345 in parallel with R2

R2345 =( R2 x R345 / R2+R345) = 26304/329 =79.95

R2345 in series with R1

79.95 + 38 =117.95 ~ 118 =Req

V= i1 x R

11.42 = i1 x 118 >>>> i1 = 0.0968A

V = E - r x i1

11.42 = 12 - r x 0.0968

r = 5.99

i1 x Req = voltage across R2 or (R 3+R4+ R5 )

0.0968 x 79.95 = i3 ( R3 +R4 + R5 ) =i3 = 0.04 A

i2 = 0.0968 - 0.04 =0.0568 A this at resistance R2

P2 = R2 x i22 = 137 x 0.05682

P2 = 0.442 watt

V2= 137 x 0.0568 = 7.782volts

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