Question

please answer my questions​

Answer 1

Question :- Find the Current in each branch of the circuit .That is $I_{AB}\:,I_{BED},\:I_{AD},\:I_{BC}\:,I_{CA}\:,I_{CD}$

Solution :-

Given

A complex electric circuit consisting of resistances and cells in different branches.

To Find

Current in each branches $I_{AB}\:,I_{BED},\:I_{AD},\:I_{BC}\:,I_{CA}\:,I_{CD}$

Concept

• This question can be solved using Kirchoff's voltage law (KVL) which states that voltage drop in a closed loop is zero.

• V = IR.

• The voltage decreases in the direction of current.

Solution

Apply KVL in branch ABCA

$V_{A}\:-10\:+I_1\:+2(I_2\:+\:I_3}\:+4I_2\:V_{A}$

=> $I_1\:+2I_2\:+2I_3\:+4I_2\:=\:10$

=> $I_1\:+6I_2\:+2I_3\:=\:10$........(I)

Apply KVL in BDEB

$V_B\:-2(I_2\:+\:I_3)\:-\:2(I_2\:+\:I_3\:-I_1)\:+5\:=V_B$

=> $-2I_2\:-\:2I_3-2I_2\:-\:2I_3\:+\:2I_2\:=\:-5$

=> $2I_2\:+\:4I_3=\:5$ ......(2)

Apply KVL in ABDA

$V_A\:-\:4I_2\:-\:2(I_2\:+\:I_3)\:-\:2(I_2\:+\:I_3\:-I_1)\:+\:4(I_1\:-\:I_2)\:=\:V_A$

=> $-12I_2\:-\:4I_3\:+\:6I_1\:=\:0$

=> $6I_1\:=\:12I_2\:+\:4I_3$

=> $I_1\:=\:\dfrac{6I_2\:+\:2I_3}{3}$

Putting this value of $I_1$ in eqn(1),we get

$\dfrac{6I_2\:+\:2I_3}{3}\:+6I_2\:+\:2I_3\:=\:10$

=> $\dfrac{6I_2\:+\:2I_3\:+\:18I_2\:+\:6I_3}{3}\:=\:10$

=> $6I_2\:+\:2I_3\:+\:18I_2\:+\:6I_3\:=\:30$

=> $12I_2\:+\:4I_3\:=\:15$.....(3)

Subtracting eqn(2) from (3) ,we get

=> $I_2\:=\:1$

Putting values of $I_2$ in equation(3)

$4I_3\:+\:12(1)\:=\:15$

=> $I_3\:=0.75$

Putting these values in eqn(1)

=> $I_1\:+6(1)\:+2(0.75)\:=\:10$

=> $I_1\:=\:2.5$

Thus, $I_1\:=\:2.5,\: I_2\:=\:1,\:I_3\:=0.75$

Current in various circuits :-

$I_{AB}\:=\:I_2\:=\:1$ A

$I_{DEB}\:=\:I_3\:=\:0.75$ A

$I_{AD}\:=\:I_1\:-\:I_2\:=1.5$ A

$I_{BC}\:=\:I_3\:+\:I_2\:=\:1.75$ A

$I_{CA}\:=\:I_1\:=\:2.5$ A

$I_{CD}\:=\:I_2\:+I_3\:-\:1\:=\:-0.75$ A