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