Question and Answer

Fig. 23(a)

Applying KVL in mesh I, II and II respectively

$- 2 I_{1} - 3 I_{2} + {7 I}_{3} = 12$ … (1)

$7 I_{1} - 4 I_{2} - 2 I_{3} = 0$ … (2)

$- 4 I_{1} + 9 I_{2} - 3 I_{3} = 0$ … (3)

$[\begin{matrix} - 2 & - 3 & 7 \\ 7 & - 4 & - 2 \\ - 4 & 9 & - 3 \end{matrix}] [\begin{matrix} I_{1} \\ I_{2} \\ I_{3} \end{matrix}] = [\begin{matrix} 12 \\ 0 \\ 0 \end{matrix}]$

Solving above equation by Cramer’s rule

$I_{1} = \frac{[\begin{matrix} 12 & - 3 & 7 \\ 0 & - 4 & - 2 \\ 0 & 9 & - 3 \end{matrix}]}{[\begin{matrix} - 2 & - 3 & 7 \\ 7 & - 4 & - 2 \\ - 4 & 9 & - 3 \end{matrix}]} = \frac{12 (12 + 18)}{182} = \frac{360}{182} = 1.978 A m p$ .

$I_{2} = \frac{[\begin{matrix} - 2 & 12 & 7 \\ 7 & 0 & - 2 \\ - 4 & 0 & - 3 \end{matrix}]}{[\begin{matrix} - 2 & - 3 & 7 \\ 7 & - 4 & - 2 \\ - 4 & 9 & - 3 \end{matrix}]} = \frac{- 12 (- 21 - 8)}{182} = \frac{348}{182} = 1.912 A m p$ .

$I_{3} = \frac{[\begin{matrix} - 2 & - 3 & 12 \\ 7 & - 4 & 0 \\ - 4 & 9 & 0 \end{matrix}]}{[\begin{matrix} - 2 & - 3 & 7 \\ 7 & - 4 & - 2 \\ - 4 & 9 & - 3 \end{matrix}]} = \frac{12 (63 - 16)}{182} = \frac{564}{182} = 3.098 A m p$ .

Hence current in different branches are 1.978 A, 1.921 A and 3.098 A.

Find the current in each branch of the network using Kirchhoff’s law

Fig.23

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Basic Electrical Engineering

Radio Frequency and Microwave Engineering

Analog Communication

Electrical Engineering Management

Power Electronics

Power System I

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Fig.22(a)

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Find the current in each branch of the network using Kirchhoff’s law Fig.23