Sum

Find the current through the 10 Ω resistor shown in the figure.

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#### Solution

Applying KVL in loop 1, we get:-

3i + 6i_{1} = 4.5 ...............(1)

Applying KVL in loop 2, we get:-

\[\left( i - i_1 \right)10 + 3 - 6 i_1 = 0\]

\[10i - 16 i_1 = - 3 ..............(2)\]

Multiplying equation (1) by 10 and (2) by 3 and then, subtracting (2) from (1), we get:-

\[- 108 i_1 = - 54\]

\[ \Rightarrow i_1 = \frac{54}{108} = \frac{1}{2} = 0 . 5\]

Substituting the value of i_{1} in (1), we get:-

\[3i + 6 \times \frac{1}{2} - 4 . 5 = 0\]

\[3i - 1 . 5 = 0\]

\[ \Rightarrow i = \frac{1 . 5}{3} = 0 . 5\]

So, current flowing through the 10 Ω resistor = i - i_{1} = 0.5 - 0.5 = 0 A

Concept: Electric Current

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