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 = 51 ?, R4 = R5 = 121 ? and R2 = 145 ?. The measured voltage across the terminals of the batery is Vbattery = 11.51 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?

- Vbattery decreases

-Vbattery increases

-Vbattery remains the same

Explanation / Answer

Potential drop across the internal resistance of the battery, Eb = 12-11.51 = 0.49 V.

Now determine the equivalent resistance of the circuit -

R3, R4 and R5 are in series. So, its equivalent = 53 + 121 + 121 = 295 Ohm

Now, this combination is in parallel with R2.

So, equivalent resistance of the combination = (145*295) / (145+295) = 97 Ohm.

Again, this is in series with R1. So, its equivalent = 97+51 = 148 Ohm.

Now, [V/(148+r)]*r = 0.49

=> [12/(148+r)]*r = 0.49

=> r/(148+r) = 0.49/12

=> 12r = 0.49(148+r)

=> 11.51r = 72.52

=> r = 6.3 ohm.

So, equivalent resistance of the combination = 148+6.3 = 154.3 Ohm

(1) I1 = 12/154.3 = 0.078 A

(2) r = 6.3 ohm.

(3) I3 = 0.078*[145/(145+295)] = 0.026 A

(4) I2 = 0.078 - 0.026 = 0.052 A

So, P2 = I2^2*R2 = 0.052^2*145 = 0.392 W

(5) V2 = I2*R2 = 0.052*145 = 7.54 V

(6) If R2 is shorted, value of the equivalent resistance of the combination increases. This means current through the internal resistance decreases. So, Vbattery increases.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.