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 = 54 ?, R4 = R5 = 98 ? and R2 = 99 ?. The measured voltage across the terminals of the batery is Vbattery = 11.76 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

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 = 54 ?, R4 = R5 = 98 ? and R2 = 99 ?. The measured voltage across the terminals of the batery is Vbattery = 11.76 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

Explanation / Answer

1)

R3 ,R4 and R5 are in series

R345=54+98+98 =250 ohms

R2 and R345 are in parallel

1/R2345 =1/99 + 1/250

R2345=70.917 ohms

R1 and R2345 are in series ,so eqyivalent resistance

Req=54+70.917 =124.917 ohms

so current I1 flowing in the circuit is

I1=Vbattery/Req=11.76/124.917

I1=0.09414 A=94.14 mA

2)

Internal Resistance

r=(E-Vbattery)/I =(12-11.76)/0.09414

r=2.55 ohms

3)

Current through R3,R4 and R5 is

I3=94.14*(99/99+250) =26.7 mA

4)

Current through R2 is

I2=I1-I3=94.14-26.7 =67.44 mA

Power dissipated in R2 is

P2=I22R2 =(67.44*10-3)2*99

P2=0.45 Watts

5)

Voltage across R2 is

V2=I2R2=(67.44*10-3)*99

V2=6.677 Volts= 6.68 Volts

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