When a voltage source is the only component in a branch that

When a voltage source is the only component in a branch that connects the reference node to another essential node, the voltage at that essential node is known. For example, the 8 V source in the circuit here is the only component in the branch that connects the node labeled v_a to the reference node, so = 8 V. This is a \"special case\"- there is no need to write a KCL equation at v_a, because the voltage is known. This special case also exists for the branch that connects the node labeled v_c and the reference node. What is the value of the voltage at v_c? Recall that v_1 = 8 V and v_2 = 17 V.

Solution

consider top loop current as i1, left side loop current a i2 and right side loop current ass i3.

all the currents are in clock wise direction.

apply kvl to loop current as i1

30i1+140(i1-i3)+25(i1-i2)=0

195i1-25i2-140i3=0

apply kvl to loop current i2

-8+25(i2-i1)+10(i2-i3)=0

-25i1+35i2-10i3=8

apply kvl to loop current i3

+17+10(i3-i2)+140(i3-i1)=0

-140i1-10i2+150i3=-17

from the above equations

i1=-0.3 amps

i2=-0.1 amps

i3=-0.4 amps

io=i1-i3=-0.3+0.4=0.1 amp.

vc=17 volts.