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In the circuit shown in Fig, 22.3, the reading ofammeter A is604Ω3Ω2Ω2 V |
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Answer» The resistance 2,3, and 6 are in parallel to each other so, 1/Req = 1/6+1/3+1/2 1/Req = (1+2+3)/6 = 6/6 = 1Ω now this is in series with 4Ω , so they add up to give Req as 4+1 = 5Ω now current will V/R = 2/5 = 0.4A reading in the ammeter is = 0.4A |
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