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

(A) Since the resistor are connected in parallel, the potential differnce across each resistor is same(i.e V=10V)
Therefor, 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_3)=(10)/(20)=0.5A`
(B) Toatl current in the circuit,
`I=I_1+I_2+I_3` (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`
(ii)Charge Q=12C, Time t=6s. Therefore,
Current `I=(Q)/(t)=(12)/(6)=2A`