In the circuit diagram given below, three resistors `R_(1),R_(2)` and `R_(3)` of `5Omega,10Omega` and `20Omega` respectively are cooncted as shown.
(A) Current through each resistor
(B) Total current in the circuit
(c ) Total resistance in the circuit

Explanation

Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e. V=10V)
Therefore, the current through `R_(1)` is
`I_(1)=(V)/(R_(1))=(10)/(5)=2A`
Current through `R_(2)=I_(2)=(V)/(R_(2))=(10)/(10)=1A`
Current through `R_(3)=I_(3)=(V)/(R)(10)/(20)=0.5A`
(B) Total current in the circuit, `I=I_(1)+I_(2)+I_(3)`
`=2+1+0.5=3.5A`
(C) Total resistance in the circuit `(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))`
`=(1)/(5)+(1)/(10)+(1)/(20)`
`=(4+2+1)/(20)`
`(1)/(R_(p))=(7)/(20)`
Hence, `R_(p)=(20)/(7)=2.857Omega`