Question

Using the figure shown, if I₃= 1.17 A, then (I₁ , I₂) A is:

a- (3.17 , 2.00)
b- (1.33 , 0.17)
c- (1.67 , 0.50)
d- (1.27 , 0.10)

Answer 1

The value of the current in each arm of the circuit is given by Kirchhoff's

Rules.

• The correct option for (I₁, I₂) is; b-. $\underline{(1.33, \ 0.17)}$

Reasons:

By Kirchhoff's junction rule, we have that the sum of current at a junction is given as follows;

• $\displaystyle \mathbf{ \sum \limits_{k = 1}^ n I_K} = 0$

Which by the direction of the currents in the given circuit diagram, we have;

• I₁ - (I₃ + I₂) = 0

Therefore;

I₁ = I₃ + I₂

According to Kirchhoff's loop rule theory, we have;

• $\displaystyle \mathbf{\sum \limits_{k = 1}^ n V_K} = 0$

In the loop having the 2 Volts emf., we have;

-I₃·R₁ + 2 + I₂·R₂ = 0

I₃ = 1.17 A

R₁ = 2 Ω

R₂ = 2 Ω

Which gives;

-1.17 × 2 + 2 + I₂×2 = 0

I₂ × 2 = 2.34 - 2 = 0.34

I₂ = 0.34 ÷ 2 = 0.17

I₂ = 0.17 A

From the loop having the 7 Volts emf. we have;

7 - 0.17×2 - I₁ × 4 - I₁ × 1 = 0

Which gives;

7 - 0.34 - 5·I₁ = 0

5·I₁ = 7 - 0.34 = 6.66

I₁ = 6.66 ÷ 5 = 1.332

Therefore;

• $\underline{\mathbf{(I_1, \, I_2)} = (1.33, \, 0.17)}$

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