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

Equivalent resistors or R of 4,6. 12 resistors connected in parallel is given by

`(1)/(R)=(1)/(4)+(1)/(6)+(1)/(12)`
Resistor of parallel combination `R=2 Omega`
Total resistor i.e., `8 Omega` is connected in series with R.
`R_(s)=8+R`
`R_(s)=8+2=10 Omega`
`:.R_(s)=10 Omega`
Net voltage (emf) `V=10V` [ `:.` cell are connected in series total emf `epsi+ epsi=2epsi`]
Circuit in through circuit `I=(V)/(R)` (from ohm's law)
`I=(10)/(10), I=1A`
So the circuit through each cell and `8Omega` resistor is 1A.
Potential drop across the parallel combination of three resistors is `V =I R'=1xx2= 2V`
`:.` Current in `4 Omega` resistor `I=(2)/(4)=0.5A[ I=(V)/(R)]`
Current in `6 Omega` resistor `I=(2)/(6)=0.33A`
Current in 12 resistor, `I=(2)/(12)=(1)/(6)=0.17A`