Question

Consider the circuit shown below. Calculate the total power dissipated by all the resistors in the circuit. R2 = 4 13 中宁 R4 = 22 V2 = 5 V V1 = 12 V Is R2 = 32 R3 = 212 12 13

Answer 1

Using kirchoff's junction law:

At junction between R2 and R3

Incoming current = Outgoing current

I1 + I3 = I2

At junction between V2 and R3

I4 = I5 + I3

Using kirchoff's Voltage law in the left loop:

V1 - I1*R1 - I2*R2 = 0

12 = 4*I1 + 3*I2

Using kirchoff's Voltage law in the right loop:

V2 - I4*R4 = 0

I4 = V2/R4 = 5/2 = 2.5 Amp

Using KVL in whole outer loop:

V1 - I4*R4 - I3*R3 - I2*R2 = 0

12 - 2.5*2 = 2*I3 + 3*I2

2*I3 + 3*I2 = 7

Now we've 4 unknown and 4 equations, So

I1 + I3 = I2

2.5 = I5 + I3

12 = 4*I1 + 3*I2

2*I3 + 3*I2 = 7

from 1st and 4th equation

2*(I2 - I1) + 3*I2 = 7

5*I2 - 2*I1 = 7

Multiply above equation with 2

10*I2 - 4*I1 = 14

4*I1 + 3*I2 = 12

Add above equations

13*I2 = 26

I2 = 2.0 Amp

I1 = (12 - 3*I2)/4 = (12 - 3*2)/4 = 1.5 Amp

I3 = I2 - I1 = 2.0 - 1.5 = 0.5 Amp

I5 = 2.5 - I3 = 2.5 - 0.5 = 2.0 Amp

Now total power dissipated by all the resistors will be:

P_total = P1 + P2 + P3 + P4 = I1^2*R1 + I2^2*R2 + I3^2*R3 + I4^2*R4

P_total = 1.5^2*4 + 2.0^2*3 + 0.5^2*2 + 2.5^2*2

P_total = 34 W

Let me know if you've any query.