Physics, asked by jyothi13, 1 year ago

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Answered by basaksarbhandar2020
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When 8 and 12 are in parallel 

When 8 and 12 are in parallel and 7.2 is in series with them 

When 8 and 12 are in parallel and 7.2 is in series with them Net Resistance 107.2+12+88×12

When 8 and 12 are in parallel and 7.2 is in series with them Net Resistance 107.2+12+88×12=1072+2096=1072+1048=10120=12Ω

When 8 and 12 are in parallel and 7.2 is in series with them Net Resistance 107.2+12+88×12=1072+2096=1072+1048=10120=12ΩTotal current =netRV=126=0.5Ω

When 8 and 12 are in parallel and 7.2 is in series with them Net Resistance 107.2+12+88×12=1072+2096=1072+1048=10120=12ΩTotal current =netRV=126=0.5ΩVoltage across R1=iR1=0.5×7.2=3.6

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