Analyze the circuit below using Loop Analysis and solve for

Analyze the circuit below using Loop Analysis and solve for all of the currents and voltages below. Write the system of equations obtained and the final answers in the boxes provided.

Solution

from the circuit IC= 3 AMPS

apply kvl to loop current IB.

-12+1(IB-IA)+14(IB-3)-4=0

-IA+15IB=58

apply kvl to loop current IC.

3IA+4(IA-3)+1(IA-IB)=0

8IA-IB=12

from the above equations

IA=2 AMPS

IB=4 AMPS

I1=IA= 2 AMPS

I2=IB-IA=4-2=2 AMPS

I3=IC-IA=3-2=1 AMPS
I4=IB-IC=4-3=1 AMPS

V1=3*IA=3*2=6 VOLTS

V2=1*I2=1*2=2 VOLTS

V3=4*I3=4*1=4 VOLTS

V4=14*I4=14*1=14 VOLTS.