The 5.00-V battery in (Figure 1) is removed from the circuit and replaced by a 1

Question

The 5.00-V battery in (Figure 1) is removed from the circuit and replaced by a 15.00 V-battery, with its terminal next to point b. The rest of the circuit is as shown in the figure. Find the current in the middle branch. Express your answer with the appropriate unite. Enter positive value if the current is to the left and negative value if the current is to the right Find the current in the lower branch Express your answer with the appropriate units Enter positive value if the current is to the left and negative value if the current is to the right Find the potential difference V_ab of point a relative to point Express your answer with the appropriate units.

Explanation / Answer

According to KVL, we have

10 - i1*(2+3) = 20-i2*(1+4) = (i1+i2) * 10

i2 = i1+2
10-i1*5 = 20*i1-20
i1 = -0.4
i2 = +1.6

So,
A) upper current 0.4A to the right
B) middle 1.6A to the left
C) bottom 1.2A to the right

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