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 = 52 , R4 = R5 = 95 and R2 = 93 . The measured voltage across the terminals of the batery is Vbattery = 11.77 V.

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

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

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

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

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

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

Explanation / Answer

Here ,

part 1)

for the resistances R3 , R4 and R5 in series ,

Rs = 52 + 95 + 95

Rs = 242 Ohm

for the equivalent resistance , Req = 52 + 242 * 93/(242 + 93)

Req = 119.2 Ohm

current in R1 , I1 = 11.77/(119.2)

I1 = 0.0987 A

I1 = 98.7 mA

the current in R1 ia 98.7 mA

part 2)

let the internal resistance is r

r = (12 - 11.77)/.0987

r = 2.33 ohm

the internal resistance is 2.33 Ohm

3)

Using current divider fornula

I3 = R2 * I1/(R2 + Rs)

I3 = 93 * 98.7 /( 93 + 242)

I3 = 27.4 mA

the current through R3 is 27.4 mA

4)

current in R2 , I2 = 98.7 - 27.4

I2 = 71.3 mA

power , P2 = I2^2 * R2

P2 = 0.0713^2 * 93

P2 = 0.473 W

the power dissipated in resistor R2 is 0.473 W

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