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: R1 = R3 = 35, R4 = R5 = 127 and R2 = 156. The measured voltage across the terminals of the batery is Vbattery = 11.69 V.

1) What is I1, the current that flows through the resistor R1?

2) What is r, the internal resistance of the battery?

3) What is I3, the current through resistor R3?

4) What is P2, the power dissipated in resistor R2?

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

6)

Resistor R2 is now shorted out as shown. How does the magnitude of the voltage across the battery change?

Vbattery decreases

Vbattery increases

Vbattery remains the same

Please use my numbers

Explanation / Answer

1.

equivalent resistance

R3 ,R4 and R5 are in series

R_345=35+127+127=289 ohms

R2 and R345 are in parallel

1/R_2345 =1/R_2 +1/R_345 =1/156 +1/289

R_2345=101.3 ohms

R1 and R_2345 are in series

Req=35+101.3 =136.3 ohms

Total current flowing in the cicrcuit is

I=V/Req=12/136.3=0.088 A or 88 mA

Current through R1 is

I1=I=0.088 A or 88 mA

b)

Internal Resistance

r=V-Vbattery/I=(12-11.69)/0.088

r=3.52 ohms

c)

current through R3 is

I3=I*(R2/R2+R345)=0.088*(156/156+289)

I3=0.03085 A or 30.85 mA

d)

Current through R2 is

I2=I-I345=0.088-0.03085 =0.05715 A

Power dissipated is

P2=I22*R2=0.057152*156=0.5095 Watts =0.51 Watts (approx)

e)

V2=I2*R2=0.05715*156 =8.915 Volts

f)

Vbattery decreases

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