In the figure R1 449 ohm R2 135 ohm and the ideal battery

In the figure R_1 = 4.49 ohm, R_2 = 13.5 ohm, and the ideal battery has emf epsilon = 12.7 V. (a) What is the magnitude of current l_1? (b) How much energy is dissipated by all four resistors in 2.96 min?

Solution

A. You can see that all three R2 resistors are in parallel.

1/Rt = 1/R2 + 1/R2 + 1/R2

Rt = 13.5/3 = 4.5 ohm

now Rt and R1 are in series

So total resistance of circuit

Req = Rt + R1 = 4.5 + 4.49 ohm = 8.99 ohm

V = i*Req

i = 12.7/8.99 = 1.412 amp

Now you know that in series circuit current is same in every resistor and in parallel circuit voltage drop is same at every circuit.

Now voltage drop on three paralle resistor will be

Vp = V - V1 = 12.7 - 4.49*1.412 = 6.36 V

Now this voltage drop will be equal on all R2 resistors because they are in parallel circuit.

i1 = i/3 = 1.412/3 = 0.470 amp.

or i1 = V/R2 = 6.36/13.5 = 0.470 amp.

B. Power = i^2*Req

P = 1.412^2*8.99 = 17.92 W

Energy = P*t = 17.92*2.96*60 = 3182.59 J

Let me know if you have any doubt.