The V = 9.10 V battery in the figure below is removed from the circuit and reins

Question

The V = 9.10 V battery in the figure below is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. (R = 9.20 ?.) (a) Find the current in each branch.

upper branch

---------- A

middle branch

---------- A

lower branch

----------A

(b) Find the potential difference Vab of point a relative to point b.

----------- V

Explanation / Answer

Taking I1 current in top loop and I2 in bottom loop both in clockwise direction Applying KVL in top loop (1+2+3+4)I1 - (1+4)I2 - 9.10 -5.00 = 0 => 10I1 -5I2 = 14.10 ........... 1 Applying KVL in bottom loop (1+4+9.20)I2 - (1+4)I1 + 5 = 0 => 14.20I2 - 5I1 = -5 ............. 2 solving 1 and 2, we get, I1 = 1.500 A I2 = 0.175 A

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