A 9 V battery has some internal resistance at room temperature. As a result, it

Question

A 9 V battery has some internal resistance at room temperature. As a result, it can provide, depending upon the load resistance R_L placed between the terminals, a maximum voltages of 9.00 V between its terminals, or a maximum current of 33.0 A. Note that in the case that it provides high current its temperature increases very quickly so its internal resistance would increase, and this condition would not last for long. Under what condition would it provide maximum voltage? Under what condition would it provide maximum current? What is the internal resistance of the battery, R, at room temperature? What is the emf of the battery, V? What power is dissipated by the internal resistance when the battery provides maximum voltage? What power is dissipated by the internal resistance when the battery provides maximum current?

Explanation / Answer

Apply KVl to the left loop

-(I1*R1)-(I2*R2)-(I1*R6)+V = 0

7*I1 + 2*I2 = 13.2 ...............(1)

Apply KVL to the middle loop

-(I3*R3)-(I3*R4)+(I2*R2) = 0

-(7*I3) + (2*I2) = 0 ................(2)

Apply KVl to the Right loop

-(I5*R5) + (I3*(R3+R4) = 0

-5*I5 + 7*I3 = 0 ...............(3)

apply KCL

I1 -I2 -I3 - I5 = 0 ........(4)

Solving (1),(2) , (3) and (4) we get

I1 = 1.61 A

I2 = 0.956 A

I3 = 0.273 A

and I5 = 0.382 A

a) V3 = I3*R3 = 0.273*3 = 0.819 V

b) I2 = 0.956 A

c) I5 = 0.382 A

and D) P = I1^2*R1 = 1.61^2*1 = 2.59 W

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