Two cells each of 5V are connected in 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.

Explanation

`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 connected in 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` = 0.17 A