The dead battery of your car provides a potential difference of 8.750 V and has

Question

The dead battery of your car provides a potential difference of 8.750 V and has an internal resistance of 1.140 O. You charge it by connecting it with jumper cables to the live battery of another car. The live battery provides a potential difference of 12.00 V and has an internal resistance of 0.0100 O, and the starter resistance is 0.0700 O.
The starter is in parallel with the live battery.

(A) Determine the current in the live battery, in the dead battery, and in the starter immediately after you closed the circuit.

Please i need help...

Explanation / Answer

Please ask if you have any doubt.I will help you.

The dead battery : 8.75V ,1.14 (series)

Live bettery : 12.00 V , 0.01(series)

Parallel resistance : 0.07

All the three are in parallel .

lets assume the current through the live battery is i and

current through the dead battery is i1 and starter resistance i2.

Now ,i = i1 + i2.

voltage across the three parallel elements is same.

Therefore , 12.00 - (0.01)i = 8.75 + (1.14)i1 = (0.07)i2

solving the above equations ,we get

i = 151.72A ,i1 = 1.52A ,i2 = 150.2A

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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