Current Electricity and Electrical InstrumentMedium

Question

Refer to the figure given below. The values of $I_{1},I_{2}$ and $I_{3}$ are $\_\_\_\_$ .

Options

A.$I_{1} = 2.5\text{ }A,I_{2} = 1.875\text{ }A,I_{3} = 1.875\text{ }A$
B.$I_{1} = 1.875\text{ }A,I_{2} = 2.5\text{ }A,I_{3} = 1.875\text{ }A$
C.$I_{1} = 1.875\text{ }A,I_{2} = 1.875\text{ }A,I_{3} = 2.5\text{ }A$
D.$I_{1} = 2.5\text{ }A,I_{2} = 2.5\text{ }A,I_{3} = 1.875\text{ }A$

Solution

Sol. [Key Concept: Apply Kirchhoff's voltage law to the three loops and solve the simultaneous equations for $I_{1},I_{2},I_{3}$.]

Loop-1 :

$$- 4\left( i_{1} - i_{3} \right) - 2\left( i_{1} + i_{2} - i_{3} \right) - i_{1} + 10 = 0$$

$$\begin{array}{r} - 7i_{1} - 2i_{2} + 6i_{3} + 10 = 0\#(1) \end{array}$$

Loop-2 :

$$- 5 - 2\left( i_{3} - i_{2} \right) + 2\left( i_{1} + i_{2} - i_{3} \right) = 0$$

$$\begin{array}{r} 2i_{1} + 4i_{2} - 4i_{3} - 5 = 0\#(2) \end{array}$$

Loop-3 :

$$- i_{1} + 10 - 4i_{3} - 2\left( i_{3} - i_{2} \right) = 0$$

$$\begin{array}{r} 2i_{2} - 6i_{3} - i_{1} + 10 = 0\#(3) \end{array}$$

On solving equation- (i) & (ii) & (iii)

$${i_{1} = 2.5\text{ }A }{i_{2} = 1.875\text{ }A }{i_{3} = 1.875\text{ }A }$$

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