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Here is a symmetrical circuit. If e pictured is 6 V and R = 8 ohms, what is the potential from a to b. This seems like a simple problem, but I can't seem to get the answer.
Thanks.
Here is a symmetrical circuit. If e pictured is 6 V and R = 8 ohms, what is the potential from a to b. This seems like a simple problem, but I can't seem to get the answer.
Thanks.
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Group resistors in right and left branches together. We'll call branches 1 (left), 2 (middle) & 3 (right).
R1 = R3 = 1Ω + 1Ω = 2Ω
KCL
+ IR1 + IR2 + IR3 = 0 (1)
Picking all currents into junction a. Incorrect currents will be negative.
KVL (left loop - ClockWise direction)
2V - 2Ω IR1 - 6V + 8Ω IR2 = 0 (2)
-4 + 8 IR2 = 2 IR1
IR1 = -2 + 4 IR2 (2A)
KVL (right loop - ClockWise direction)
-8Ω IR2 + 6V + 2Ω IR3 - 2V = 0 (3)
2 IR3 = -4 + 8 IR2
IR3 = -2 + 4 IR2 (3A)
Substitute 2A and 3A into 1.
+ IR1 + IR2 + IR3 = 0
( -2 + 4 IR2 ) + IR2 + ( -2 + 4 IR2 ) = 0
9 IR2 = 4
IR2 = 0.444A
Solve 2A and 3A for other currents.
IR1 = -2 + 4 IR2 = -2 + 4 * 0.444 = -0.222A
IR3 = -2 + 4 IR2 = -2 + 4 * 0.444 = -0.222A
Vab = 2 V - IR1 2 R = 2 V - (-0.222A) * 2Ω = 2.444V
R1 = R3 = 1Ω + 1Ω = 2Ω
KCL
+ IR1 + IR2 + IR3 = 0 (1)
Picking all currents into junction a. Incorrect currents will be negative.
KVL (left loop - ClockWise direction)
2V - 2Ω IR1 - 6V + 8Ω IR2 = 0 (2)
-4 + 8 IR2 = 2 IR1
IR1 = -2 + 4 IR2 (2A)
KVL (right loop - ClockWise direction)
-8Ω IR2 + 6V + 2Ω IR3 - 2V = 0 (3)
2 IR3 = -4 + 8 IR2
IR3 = -2 + 4 IR2 (3A)
Substitute 2A and 3A into 1.
+ IR1 + IR2 + IR3 = 0
( -2 + 4 IR2 ) + IR2 + ( -2 + 4 IR2 ) = 0
9 IR2 = 4
IR2 = 0.444A
Solve 2A and 3A for other currents.
IR1 = -2 + 4 IR2 = -2 + 4 * 0.444 = -0.222A
IR3 = -2 + 4 IR2 = -2 + 4 * 0.444 = -0.222A
Vab = 2 V - IR1 2 R = 2 V - (-0.222A) * 2Ω = 2.444V