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Question

Find the value of i1/i2 in the figure (32-E3) if (a) R = 0.1 Ω (b) R = 1 Ω and (c) R = 10 Ω. Note from your answers that in order to get more current from a combination of two batteries, they should be joined in parallel if the external resistance is small and in series if the external resistance is large, compared to the internal resistance.

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Solution

(a) For R = 0.1 Ω:
Applying KVL in the given circuit, we get:

0.1i1+1i1-6+1i1-6=00.1i1+1i1+1i1=12i1=122.1=5.71 A
Now, consider the given circuit.

Applying KVL in the loop ABCDA, we get:

0.1i2+1i-6=00.1i2+i=6i=6-0.1i2

Applying KVL in ADEFA, we get:

i-6+6-i2-i1=0i-i2+i=02i-i2=026-0.1 i2-i2=0 i2=10 A

i1i2=0.571

(b) For R = 1 Ω:
Applying KVL in the circuit given in figure 1, we get:
1i1+1.i1-6+i1-6=03i1=12i1=4
Now, for figure 2:
Applying KVL in ABCDA, we get:
i2+i-6=0 i2+i=6
Applying KVL in ADEFA, we get:
i-6+6-i2-i1=0i-i2+i=02i-i2=026-i2-i2=0 12-3i2=0i2=4 Ai1i2=1

(c) For R = 10 Ω:
Applying KVL in the circuit given in figure 1, we get:
10i1+1i1-6+1i1-6=0
12i1=12i1=1
Now, for figure 2:
Applying KVL in ABCDA, we get:
10i2+i-6=0i=6-10i2
Applying KVL in ADEFA, we get:
i-6+6-i2-i1=0i-i2+i=02i-i2=026-10i2-i2=0 12-21i2=0i2=0.57 Ai1i2=1.75

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