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R2 R1 C R2 D
Expert Answer
By symmetry, in the circuit, we can say that the current in the equals resistance are the same. In R1 there are a current of I1 , in R2 there are a current of I2 and in R3 there are I3.
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A (2) C 2 (1) -ww wr ()
Now, from the circuit we can find the currents :
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At node B, applying KCL we get I I -I2 --- Eq. (1) Applying KVL in the outer loop (XABDY), we get E I,R2 I,R, 14.2 I,(3490) + I,(1290) --- Eq. (2) Applying KVL in the imer loop (ABC), we get LR2 R IR RI R R1 R 3 (from Eq. 1) ---- Eq. (3) R2 R3
Further,...
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Putting Eq 3 in Eq 2, we get 14.2 1,(3490)+ I,(1290) R2 R3 (1290 3650) ( 3490) + I,(1290) 14.2 (34903650) I 3.833 mA Therefore, from Eq 3 we get I, = 2.652 mA and, I 1.181 mA
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