Kirchhoff's loop rule:
V = iR
Let's apply the Kirchhoff's loop rule starting with the 8V battery going in the clockwise direction.
8 - i0.5 - i8 - i6 - i0.5 - 4 - i9 = 0
- i24 = - 8
i = 1/6 A
Starting from point a to point b, we have:
Va + i8 + i0.5 - 8 - Vd = 0
a) What is the potential difference Vad in thecircuit?
b) What is the terminal voltage of the 4 V battery?
c) A battery with emf 10.3 V and internal resistance .5ohm isinserted in the circuit d with its negative terminal connected tothe negative terminal of the 8 V battery. WHat is the difference ofpotential Vbc between the terminals of the 4 V batterynow?
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