Question

Using Kirchhoff’s rules calculate the current through the 40 Ω and 20 Ω resistors in the following circuit:

Answer 1

Kirchhoff’s junction rule: The sum of currents entering a junction is equal to the sum of currents leaving it.

Kirchhoff’s loop rule: The sum of potential differences around a closed loop is zero

Let us label the circuit with currents in accordance with Kirchhoff’s junction rule.

Using Kirchhoff’s loop rule on ABCDA,

80 - 20I₁ - 40I₂ = 0

8 - 2I₁ - 4I₂ = 0

I₁ + 2I₂ - 4 = 0

I₁ + 2I₂ - 4 = 0     ......(1)

Using Kirchhoff’s loop rule on CDEFC,

-40I₂ - 40 + 10(I₁ - I₂) = 0

-40I₂ - 40 + 10I₁ - 10I₂ = 0

-50I₂ - 40 + 10I₁ = 0

I₁ - 5I₂ - 4 = 0     ......(2)

Subtracting (2) from (1), we get

7I₂ = 0

I₂ = 0

Substituting the value of I₂ in (1), we get

I₁ = 4A

Hence, there is no current through the 40Ω resistor and 4A current flows through the 20Ω resistor.

Answer 2

Kirchhoff’s rules are as follows:

(i) Junction rule (or current rule): The sum of all currents entering a junction is equal to sum of all currents leaving the junction.

(ii) Voltage rule: The algebraic sum of changes in the potential around any closed loop must be zero.

Apply Kirchhoff’s voltage rule 2 in loop ABCDA:

Apply Kirchhoff’s voltage rule 2 in loop FEDCF:

Solving Eqs. (1) and (2) [and multiplying Eq. (1) by 4 and Eq. (2) by 3 and add], we get

Substituting i₂ = 4 in Eq. (1), we get

Therefore, the current passing through 20 Ω resistor is 4 A and the current passing through 40 Ω resistor is

i₁ – i₂ = 4 A – 4 A = 0 A