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Using Kirchhoffs rules in the given circuit, determine (i) the voltage drop across the unknown resistor R and (ii) current I in the arm EF. |
Answer» Solution :Taking upper PORTION as mesh using Kirchhoffs second law portion  `1 xx 2 -I xx 3 = 3 -5` or ` 2 - 3 I =-2` or ` 3 I =4` or `I = 4/3 A` So total current through R, ` I . =1 + 4/3 + 7/5 A` `therefore` VOLTAGE drop across R, `5-3 xx 4/3 =1` Volt. |
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