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Question
Find the circuit in the three resistors shown in the figure.
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Solution
Applying KVL in loop 1, we get:-
\[2 + \left( i_1 - i_2 \right) \times 1 - 2 = 0\]
\[ \Rightarrow i_1 = i_2\]
Applying KVL in loop 2, we get:-
\[2 + \left( i_2 - i_3 \right) \times 1 - 2 - \left( i_1 - i_2 \right) \times 1 = 0\]
\[ \Rightarrow i_2 - i_3 - i_1 + i_2 = 0\]
\[ i_1 = i_2 \]
\[ \Rightarrow i_2 - i_3 - i_2 + i_2 = 0\]
\[ \Rightarrow i_2 = i_3\]
Applying KVL in loop 3, we get:-
\[2 + i_3 - 2 - \left( i_2 - i_3 \right) = 0\]
\[ \Rightarrow i_3 = 0\]
\[ i_1 = i_2 = i_3 \]
\[ \therefore i_1 = i_2 = i_3 = 0\]
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