# In the circuit given below, R = 15 Ω. Find vo using mesh analysis.?

Relevance
• 1 year ago

To write mesh equations add all the resistance in a loop, change to negative

giving -30i1. The i2 resistance is the resistor common to loops 1 and 2 which is 20 in this case. Assume clockwise currents i1 and i2 in left and right loops. Equation for left loop:

-30i1+20i2-60=0 (-60 because that is a voltage drop in clockwise direction).

20i2 = + 30i1 + 60

i2 = (1.5i1 + 3)

Notice the dependent source 5vo = 5*(-4i1) = -20i1

Equation of the right loop

20i1 - 35i2 + 60 - 5Vo = 0 but 5Vo = -20i1 and -5Vo = 20i1

Giving:

20i1 - 35i2 +60 +20i1 = 0

But i2 = (1.5i1 + 3) from above

Substitute in

20i1 -35(1.5i1 + 3) + 60 + 20i1 = 0

20i1 -52.5i1 -105 + 60 + 20i1 = 0

-12.5i1 - 45 = 0-----> i1 = 45/(-12.5) = -3.6

Vo = -4i1 = -4*(-3.6) = 14.4Volts

i2 = 1.5(-3.6) + 3 = -2.4

Verify by comparing power IN from sources to power consumed by resistance:

10Ω*3.6²+20Ω*1.2²+15Ω*2.4² = 244.8W by resistance

60V*1.2A+72V*2.4A = 244.8W from sources Checks

i1 = 3.6A counter clock wise

i2 = 2.4 counter clock wise

i1-i2 = 1.2A up the middle branch

Do Node solution real quick. Lev V be node above +60V source. Write KCL of currents leaving V assuming bottom of ckt is ground

V/10+(V-60)/20+(V-5Vo)/15 = 0

But Vo = V/10 *4 = 0.4V so 5Vo = 2V

V/10+(V-60)/20+(V-2V)/15 = 0

Multiply equation by 60

6V+3V-180+4V-8V= 0

5V = 180 so V = 180/5 = 36V

i1 = V/10 = 36/10 = 3.6 down into 6Ω

(36-60)/20 = -1.2 or 1.2A up out of the 60V source

(V-5Vo)/15 = -V/15 = -36/15 = -2.4 or 2.4 up out of 5Vo source

• Roger
Lv 7
1 year ago

Vo = 14.4 volts

First Mesh equation ( the left loop)

4 (I1) + 6(I1) + 20( I1-I2) + 60 volts = 0

note the + end of the 4 ohm resistor is on the bottom, the + end of the 6 ohm resistor is on the left, and the + end of the 20 ohm resistor is on the top

30(I1) - 20(I2) = -60 volts

divide by 10

3I1 - 2I2 = -6

second mesh equation ( right loop)

(I2)15 +5vo -20(I1 -I2)- 60V =0

but vo = -4I1

15(I2)-20(I1) -20(I1 -I2) = 60 volts

60 volts = -40(I1) +35(I2)

divide by 5

12 volts = -8I1 +7I2

3I1 - 2I2 = -6 volts

solve for I1 and I2

I1 = -3.6 amps

I2 = - 2.4 amos

Vo = - 4(-3.6 amps) = 14.4 volts