Using Kirchhoff's laws of current and voltage, find the current in 2 Ω resistor in the given circuit.

Solution:

Using Kirchhoff's junction rule,
At junction a,

I_{1} + I_{3} = I_{2} ... (i)

Using Kirchhoff's loop rule,
In loop ecdfe,

\begin{aligned} - I_{1} \times R_{1} + I_{3} \times R_{3} - E_{2} + E_{1} & = 0 \\ - I_{1} \times 3 + I_{3} \times 4 - 40 + 35 & = 0 \\ 4 I_{3} - 3 I_{1} & = 5 ... (ii) \end{aligned}

In loop eabfe,

\begin{aligned} - I_{1} \times R_{1} - I_{2} \times R_{2} + E_{1} & = 0 \\ - I_{1} \times 3 - I_{2} \times 2 + 35 & = 0 \\ - 3 I_{1} - 2 (I_{1} + I_{3}) + 35 & = 0 \\ - 3 I_{1} - 2 I_{1} - 2 I_{3} + 35 & = 0 \\ - 5 I_{1} - 2 I_{3} + 35 & = 0 \\ 5 I_{1} + 2 I_{3} & = 35 ... (iii) \end{aligned}

Multiplying equation (iii) by 2 and subtracting (ii) from (iii), we get,

\begin{aligned} 4 I_{3} + 10 I_{1} - 70 - (4 I_{3} - 3 I_{1} - 5) & = 0 \\ 13 I_{1} & = 65 \\ ∴ I_{1} & = 5 A \end{aligned}

Substituting this in equation (ii),

\begin{aligned} 4 I_{3} - 3 \times 5 & = 5 \\ ∴ I_{3} & = 5 A \end{aligned}

Thus, from equation (i),

I_{2} = I_{1} + I_{3} = 10 A

Hence, the current through 2 Ω resistor is 10 A .

Physics in Depth