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 V in series with an internal resistance r as shown above left. The values for the resistors are: R_1 = R_3 = 58 Ohm, R_4 = R_5 = 77 Ohm and R_2 = 137 Ohm. The measured voltage across the terminals of the batery is V_battery = 11.62 V. What is l_1, the current that flows through the resistor R_1? What is r, the internal resistance of the battery? What is l_3, the current through resistor R_3? What is P_2, the power dissipated in resistor R_2? What is V_2, the magnitude of the voltage across the resistor R_2? Resistor R_2 is now shorted out as shown. How does the magnitude of the voltage across the battery change?

Explanation / Answer

Requi = R1 + ( R2 || R3+R4+R5) = 58 + ( 137 || 58+77+77) = 58 + ( 137 || 212) = 58 + 83.22 = 141.22 ohm

a) I1 = V/Requi = 12.0/141.22 = 85 mA

b) By KVL,

V - I1R1 – I2R2 – I2r = 0

12.0 - 85*10^-3*58 – I2*137 – (12.0-11.62) = 0   = > I2= 48mA

v= I2*r

12-11.62 = 48*10^-3*r    => r = 7.9 ohm

c) I3= I1 – I2 = (85-48)mA = 37mA

d) P2 = I2^2*R2 = (48*10^-3)^2*137 = 0.316 W

e)V2= I1*R2 = (48*10^-3)*137 = 6.6 V

f) Vbattery remains same , because I1 flowing through the battery is as before.

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