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 = 12V in series with an internal resistance r as shown above left. The values for the resistors are: R_1 = R_3 = 29 ohm, R_4 = R_5 = 92 ohm and R_2 = 86 ohm. The measured voltage across the terminals of the battery is V_batter4y = 11.58 V. What is I_1, the current that flows through the resistor R_1? What is r, the internal resistance of the battery? What is I_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

1.
R3,R4 and R5 are in series.
their equivalent resistance be R'
R' = R3+R4+R5
= 29+92+92
= 213 ohm

R' and R2 are in parallel
their equivalent resistance be R''
R'' = (R'*R2)/(R'+R2)
= (213*86)/(213+86)
= 61.26 ohm

This R'' and R1 are in series and coltage across them is 11.58 V

I through R1 = Inet = V/Rnet
= 11.58/(61.26+29)
= 0.128 A
Answer: 128 A

2.
across battery
Vbattery = E-Inet*r
11.58 = 12 - 0.128*r
r = 3.28 ohm
Answer: 3.28 ohm

3.
V across R2 = V across R'' = vbattery* R'' / (R'' + R1)
= 11.58*61.26 / (61.26+29)
= 7.86 V
I2 = V2/R2
= 7.86/86
= 0.091 A
= 91 A
Answer: 91 A

4.
P2 = V2^2/R2
= 7.86^2 / 86
= 0.718 A
= 718 mA

I am allowed to answer only 4 parts at a time

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