Question

Calculate i in the given circuit applying
Kirchhoff's laws.
​

Answer 1

In order to find current in more than one branch, we use kirchoff's voltage law (KVL) which is also called as kirchoff's second law.

• It states that, for a closed loop series path the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero.

Mathematically, ∑ε = ∑IR

★ Loop - 1 [ABCDA] :

➙ 6 - 6i₁ - 2i = 0

➙ 6 - 2i = 6i₁

➙ 3 - i = 3i₁

➙ i₁ = $\sf\dfrac{3-i}{3}$ $\dots$ (I)

★ Loop - 2 [AEFDA] :

➙ 6 - 12i₂ - 2i = 0

➙ 3 - 6i₂ - i = 0

➙ 3 - i = 6i₂

➙ i₂ = $\sf\dfrac{3-i}{6}$ $\dots$ (II)

We know that :: i = i₁ + i₂ $\dots$ (III)

$\sf:\implies\:i=\dfrac{3-i}{3}+\dfrac{3-i}{6}$

$\sf:\implies\:i=\dfrac{6-2i+3-i}{6}$

$\sf:\implies\:i=\dfrac{9-3i}{6}$

$\sf:\implies\:6i=9-3i$

$\sf:\implies\:2i=3-i$

$\sf:\implies\:i=\dfrac{3}{3}$

$:\implies\:\underline{\boxed{\bf{\green{i=1\:A}}}}$

• From the first equation,

$\sf:\implies\:i_1=\dfrac{3-i}{3}$

$\sf:\implies\:i_1=\dfrac{3-1}{3}$

$\sf:\implies\:i_1=\dfrac{2}{3}$

$:\implies\:\underline{\boxed{\bf{\gray{i_1=0.67\:A}}}}$

• From the third equation,

$\sf:\implies\:i=i_1+i_2$

$\sf:\implies\:1=0.67+i_2$

$\sf:\implies\:i_2=1-0.67$

$:\implies\:\underline{\boxed{\bf{\orange{i_2=0.33\:A}}}}$

Answer 2

To find:

Value of current 'i' using Kirchhoff's Law?

Calculation:

• Here we will take certain loops and use Kirchhoff's Voltage Law.

In the smaller loop :

$\rm \therefore \: - 6i_{1} + 12i_{2} = 0$

$\rm \implies \: 6i_{1} = 12i_{2}$

$\rm \implies \: i_{1} = 2i_{2}$

In the larger loop :

$\rm \therefore \: - 12i_{2} - 2i + 6 = 0$

$\rm \implies \: 12i_{2} + 2i = 6$

$\rm \implies \: 12i_{2} + 2(i_{1} + i_{2})= 6$

$\rm \implies \: 12i_{2} + 2(2i_{2} + i_{2})= 6$

$\rm \implies \: 12i_{2} + 2(3i_{2} )= 6$

$\rm \implies \: 18i_{2} = 6$

$\boxed{ \rm \implies \: i_{2} = \dfrac{1}{3} \: amp}$

$\boxed{ \rm \implies \: i_{1} =2 \times i_{2} = \dfrac{2}{3} \: amp}$

$\boxed{ \rm \implies \:i = i_{1} + i_{2} = 1 \: amp}$

Hope It Helps.