Question.
In the circuit below, R1 and R2 are in parallel and R3 is in series with this combination. There is no immediate place to calculate a voltage drop by Ohm's Law because none of the resistors has both a known current and resistance. Neither can the total resistance be obtained because R2 is unknown. So how can to find the current in R3 by applying Kirchhoff's current Law to node A?
Answer.
Let current in R1 = i and in R3 = i'
Applying Kirchoff's first law to node A, i + .001 = i'
Applying Kirchhoff's second law to the outermost loop,
76 = 56000i + 56000 ( i + 0.001)
=> i = 20 /112000 = 0.00018 A = 0.18 mA
=> i' = 1.18 mA
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