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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)

★ Loop - 2 (CDEFC)

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