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Two cells each of 5V are connectedin series across a 8 Omega resistor and three parallel resistors of 4Omega, 6 Omega and 12 Omega. Draw a circuit diagram for the above arrangement. Calculate (i) the current drawn fron the cell (ii) current through each resistor. |
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Answer» Solution :`V_1=5V , V_2=5V` `R_1=8Omega, R_2=4Omega , R_3=6Omega , R_4=12Omega` THREE resistors `R_2,R_3` and `R_4` are connected PARALLEL combination `1/R_p=1/R_2+1/R_3+1/R_4` `=1/4+1/6+1/12=3/12+2/12+1/12=6/12` `R_p=2Omega`  Resistors `R_1` and `R_p` are connectedin series combination `R_s=R_1+R_p` =8+2=10 `R_s=10Omega` TOTAL voltage connected series to the circuit `V=V_1+V_2` =5+5=10 V=10 V (i) Current through the circuit , `I=V/R_s=10/10` I=1 A Potential drop across the parallel combination , `V.=1 R_p=1xx2` V.=2V (ii) Current in `4Omega` resistor , `I=(V.)/R_2=2/4`=0.5 A Current in `6Omega` resistor , `I=(V.)/R_3=2/6`=0.33 A Current in `12Omega` resistor , `I=(V.)/R_4 =2/12` = 0.17A |
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