Question

For the given network, find
the current through 4 ohm and 3 ohm. Assume that the cells have negligible internal resistance.​

Answer 1

Hey mate! You forgot to attach diagram. Always make sure to attach diagram as reference.

Diagram : Refer to the attached image.

To Find :

Current through 4Ω and 3Ω resistor.

Solution :

❒ Current flow in simple resistor circuits can be calculated by using ohm's law simply.

In order to find current in more than one branch, we will 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

Assume that,

• current flow in 3Ω (CF) = I₁
• current flow in 4Ω (CD) = I₂
• current flow in 4Ω (AF) = I

Net current flow, I = I₁ + I₂

★ Loop - 1 (ABCFA)

$\sf:\implies\:5-3I_1-4I=0\:\dots\:(1)$

$\sf:\implies\:5-3I_1-4(I_1+I_2)=0$

$\sf:\implies\:5-3I_1-4I_1-4I_2=0$

$\sf:\implies\:5-7I_1-4I_2=0\:\dots\:(1)$

★ Loop - 2 (CDEFC)

$\sf:\implies\:10+3I_1-4I_2=0\:\dots\:(2)$

On solving both equations,

➠ 5 - 7I₁ = 10 + 3I₁

➠ 10I₁ = -5

➠ I₁ = -1/2

➠ I₁ = -0.5A (F → C)

[Negative sign indicates opposite direction]

From equation (1),

➠ 5 - 7I₁ - 4I₂ = 0

➠ 5 - 7(-0.5) - 4I₂ = 0

➠ 4I₂ = 5 + 3.5

➠ I₂ = 8.5/4

➠ I₂ = 2.121A

Current flow in 4Ω (b/w AF)

➛ I = I₁ + I₂

➛ I = -0.5 + 2.121

➛ I = 1.621 A

Answer 2

Applying Kirchhoff first lawAt junction F,

I I I 1 3 = 2 – 1 2 . 3 --- (1)

Applying Kirchhoff second law,

(i) loop EFCDE,

3 4 10 0 2 1

4 3 10 1 2 I --- (2)

(ii) loop FABCF

4 3 5 0 3 2 I I

4 3 5 3 2 I I --- (3)

From Eq. (1) and Eq. (2)

4 3 3 2 2 = 10

3 4 4 10 2 3 2

4 7 10 3 2 I I --- (4)

From Eq. (3) and Eq. (4)

10 5 2 I

I A 2 . 0 5

Negative sign indicates that I 2 current

flows from F to C

From Eq. (2) 4 3 0 5 10 1I .

I A 1 = 2 1. 2

∴ I I 3 1 I A