In the circuit shown in the figure below, R = 9.0 , 1 = 11.0 V and 2 = 7.0 V. (a

Question

In the circuit shown in the figure below, R = 9.0 , 1 = 11.0 V and 2 = 7.0 V.

(a) Find the current through the 9.0 resistor.
A

(b) Find the total rate of dissipation of electrical energy in the 9.0 resistor and in the internal resistance of the batteries.
W
(c) In one of the batteries, chemical energy is being converted into electrical energy. In which one is this happening?

12    

At what rate?
W
(d) In one of the batteries, electrical energy is being converted into chemical energy. In which one is this happening?

12    

At what rate?
W

(e) Show that the overall rate of production of electrical energy equals the overall rate of consumption of electrical energy in the circuit.

Explanation / Answer

given that

R = 9 ohm

E1 = 11 V

E2 = 7 V

r1 =1ohm

r2 =1 ohm

part(a)

apply KVL for this loop

Let clockwise current is i.

-11 - i*1 - i*1 + 7 - i*9 = 0
-4V - i*11 = 0

i*11 = -4

i = - 0.36A

So i = 0.36 A

negetive sign shows that the direction we took, is opposite to the actual direction.

part(b)

P = i^2 *R

P = (0.36)^2 *9 = 1.17 W

both 1 ohm resistors consume

i^2*1 = 0.36*0.36 *1 = 0.129 W

patr (c)

The 11 V is outputting electrical power = V*i = 11*0.36 = 3.96 W

part(d)

The 7V battery is consuming power at the rate of 7*0.36 = 2.52 W as can be seen by the fact the current i is entering at the positive terminal .

part(e)

The Output Power of the 11 V supply is 11*0.36 = 3.96 W
2.52 W are being consumed by the 7V battery.
The 9 ohm resistor is consuming 1.17 W as shown above and each 1 resistor is consuming 0.129 W
3.96 = 2.52 + 1.17 + 0.129 + 0.129

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