State Kirchhoff’s rules of current distribution in an electrical network. Using these rules determine the value of the current I₁ in the electric circuit given below.

Solution

For electrical network, Kirchhoff’s rules are as follows:

(i) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction
∴ Σ I = 0

(ii) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.
∴ Σ IR + Σ E = 0

According to Kirchhoff’s rule, I₁ + I₂ = I₃

In the given circuit,
Applying loop rule to both the lower and upper loops, we get

40 I₃ + 20 I₁ = 40 (In loop ABCF)
40 I₃ + 20 I₂ = 80 + 40 (In loop CDEF)

Adding these two equations, we get
$80 I_{3} + 20 (I_{1} + I_{2}) = 160$

$80 I_{3} + 20 I_{3} = 160$

   $I_{3} = \frac{160}{100} = 1.6 A$

Putting the value of I₃ in the above equation to find the value of I₁, we get

   $40 \times 1.6 + 20 I_{1} = 40$

$20 I_{1} = 40 - 64 = - 24$

   $I_{1} = - \frac{24}{20} = - 1.2 A .$

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State Kirchhoff’s rules of current distribution in an electrical network. Using these rules determine the value of the current I₁ in the electric circuit given below.

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Current Electricity

State Kirchhoff’s rules of current distribution in an electrical network. Using these rules determine the value of the current I1 in the electric circuit given below.

Some More Questions From Current Electricity Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

State Kirchhoff’s rules of current distribution in an electrical network. Using these rules determine the value of the current I₁ in the electric circuit given below.