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Question
Find the value of i1/i2 in the following figure 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:-
\[0 . 1 i_1 + 1 i_1 - 6 + 1 i_1 - 6 = 0\]
\[ \Rightarrow 0 . 1 i_1 + 1 i_1 + 1 i_1 = 12\]
\[ \Rightarrow i_1 = \frac{12}{\left( 2 . 1 \right)} = 5 . 71 A\]
Now, consider the given circuit.
Applying KVL in the loop ABCDA, we get:-
\[0 . 1 i_2 + 1i - 6 = 0\]
\[ \Rightarrow 0 . 1 i_2 + i = 6\]
\[ \Rightarrow i = 6 - 0 . 1 i_2\]
Applying KVL in ADEFA, we get:-
\[i - 6 + 6 - \left( i_2 - i \right)1 = 0\]
\[ \Rightarrow i - i_2 + i = 0\]
\[ \Rightarrow 2i - i_2 = 0\]
\[ \Rightarrow 2\left[ 6 - 0 . 1 i_2 \right] - i_2 = 0\]
\[ \Rightarrow i_2 = 10 A\]
\[\therefore \frac{i_1}{i_2} = 0 . 571\]
(b) For R = 1 Ω
Applying KVL in the circuit given in figure 1, we get:-
\[1 i_1 + 1 . i_1 - 6 + i_1 - 6 = 0\]
\[ \Rightarrow 3 i_1 = 12\]
\[ \Rightarrow i_1 = 4\]
Now, for figure 2:-
Applying KVL in ABCDA, we get:-
\[i_2 + i - 6 = 0\]
\[ \Rightarrow i_2 + i = 6\]
Applying KVL in ADEFA, we get:-
\[i - 6 + 6 - \left( i_2 - i \right)1 = 0\]
\[ \Rightarrow i - i_2 + i = 0\]
\[ \Rightarrow 2i - i_2 = 0\]
\[ \Rightarrow 2\left[ 6 - i_2 \right] - i_2 = 0\]
\[ \Rightarrow 12 - 3 i_2 = 0\]
\[ \Rightarrow i_2 = 4 A\]
\[ \therefore \frac{i_1}{i_2} = 1\]
(c) For R = 10 Ω
Applying KVL in the circuit given in figure 1, we get:-
\[10 i_1 + 1 i_1 - 6 + 1 i_1 - 6 = 0\]
\[\Rightarrow 12 i_1 = 12\]
\[ \Rightarrow i_1 = 1\]
Now, for figure 2:-
Applying KVL in ABCDA, we get:-
\[10 i_2 + i - 6 = 0\]
\[ \Rightarrow i = 6 - 10 i_2\]
Applying KVL in ADEFA, we get:-
\[i - 6 + 6 - \left( i_2 - i \right)1 = 0\]
\[ \Rightarrow i - i_2 + i = 0\]
\[ \Rightarrow 2i - i_2 = 0\]
\[ \Rightarrow 2\left[ 6 - 10 i_2 \right] - i_2 = 0\]
\[ \Rightarrow 12 - 21 i_2 = 0\]
\[ \Rightarrow i_2 = 0 . 57 A\]
\[ \therefore \frac{i_1}{i_2} = 1 . 75\]
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