Questions: For the circuit in the given figure, use KVL to find the branch voltages V1 to V4. Assume A=9 V, B=10 V, C=3 V, and D=11 V. The branch voltages V1, V2, V3, and V4 are V, V, V, and V, respectively.

For the circuit in the given figure, use KVL to find the branch voltages V1 to V4. Assume A=9 V, B=10 V, C=3 V, and D=11 V.

The branch voltages V1, V2, V3, and V4 are V, V, V, and V, respectively.

Transcript text: For the circuit in the given figure, use $K V L$ to find the branch voltages $V_{1}$ to $V_{4}$. Assume $A=9 V, B=10 V, C=3 V$, and $D=11 V$. The branch voltages $V_{1}, V_{2}, V_{3}$, and $V_{4}$ are $\square$ V, $\square$ V, $\square$ V , and $\square$ V, respectively.

Solution

Solution Steps

Step 1: Applying KVL to the left loop

Going clockwise around the left loop, we can write the KVL equation:

-A + V1 + V3 - B = 0

Substituting the given values A = 9V and B = 10V:

-9 + V1 + V3 - 10 = 0 V1 + V3 = 19

Step 2: Applying KVL to the right loop

Going clockwise around the right loop, we can write the KVL equation:

-V3 + C + V2 - D = 0

Substituting the given values C = 3V and D = 11V:

-V3 + 3 + V2 - 11 = 0 V2 - V3 = 8

Step 3: Solving for V1, V2, V3, and V4

We have two equations:

(1) V1 + V3 = 19 (2) V2 - V3 = 8

Also note that due to the parallel configuration we have:

(3) V1 = V4

Adding equations (1) and (2): V1 + V2 = 27

From the figure, we notice that V3 and C are parallel branches, meaning that V3 = C. Therefore:

V3 = 3V

Substituting V3 = 3V into equation (1): V1 + 3 = 19 V1 = 16V

Substituting V3 = 3V into equation (2): V2 - 3 = 8 V2 = 11V

Since V1 = V4 V4 = 16V

Final Answer:

The branch voltages V1, V2, V3, and V4 are 16 V, 11 V, 3 V, and 16 V, respectively.

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >

Thank you for your feedback.

Copied!