1.

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

Answer»

For loop ABCDA, 

20I + 40(I – I1) = 80 

⇒ 60I – 40 I1 = 80 

⇒ 3I – 2I1 = 4 ……….. (1) 

For loop EDCFE,

-40(I – I1) + 10I1 = 40 

-40 I + 40I1 + 10I1 = 40 

50I1 – 40I = 40 

5I2 – 4I = 4 …………. (2) 

From (1) and (2),

On multiplying (1) × 4 and (2) × 3 we get; 

15I1 – 8I1 = 28 

7I1 = 28 

I1 = 4A 

Putting in (1), 

3I - 2× 4 = 4 

3I = 12 

I = 4A 

∴ Current through 20Ω is I = 4A 

Current through 40Ω in (I – I1) = 4 – 4 = 0



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