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=10V`
(i) Current through the circuit, `I=(V)/(R_(s))=(10)/(10)`
`I=1A`
Potential drop across the parallel combination,
`V.=IR_(p)=1xx2`
`V.=2V`
(ii) Current in `4Omega` resistor, `I=(V.)/(R_(2))=(2)/(4)=0.5A`
Current in `6Omega` resistor, `I=(V.)/(R_(3))=(2)/(6)=0.33A`
Current in `12Omega` resistor, `I=(V.)/(R_(4))=(2)/(12)=0.17A`