A circuit is constructed with five resistors and one real battery as shown above

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 Vin series with an internal resistance r as shown above left. The values for the resistors are: R1 R3 42 n, R Rs 82 n and R2 147 o. The measured voltage across the terminals of the batery is vbattery 11.8 v. Ia Ra II R I, Ra 1) What is l1, the current that flows through the resistor R1? 92.46 mA Submit 2) What is r, the internal resistance of the battery? 2.166 Submit http://www fl pit physics.com/Course Fainproblem

Explanation / Answer

V1 = voltage across R1 = i1 R1 = (0.09246) (42) = 3.88 volts

V2 = Voltage across R2 = Vbattery - V1 = 11.8 - 3.88 = 7.92 volts

i2 = current in R2 = V2/R2 = 7.92/147 = 0.05388 A

from the circuit , the current i1 divides into i2 and i3 hence

i1 = i2 + i3

0.09246 = 0.05388 + i3

i3 = 0.03858 A = 38.6 mA

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