In the circuit diagram given below, three resistors `R_(1),R_(2)`, and `R_(3)` of `5 Omega, 10 Omega` and `30 Omega`, respectively are connected as shown.

Calculate:
(a) current through each resistor.
(b) total current in the circuit.
(c) total resistance in the circuit.

Explanation

`R_(1)=5 Omega, R_(2)=10Omega, " and " R_(3)=30 Omega`.
Potential difference across the battery, V=12 V.
This is also the potential difference across each of the individual resistor, therefore, to calculate the current in the resistores, we use Ohm's law.
The current `I_(1),"through"R_(1)=V//R_(1)`
`I_(1)=12C//5Omega=2.4 A`.
`"The current" I_(2),"thrgough"R_(2)=V//R_(2)`
`I_(2)=12V//10Omega=1.2 A.`
`" The currnt "I_(3), " through " R_(3)=V//R_(3)`
`I_(3)=12V//30Omega=0.4 A.`
The total currnt in the circuit.
`I=I_(1)+I_(2)+I_(3)`
`=(2.4+1.2+0.4)A`
`=4A`
`"The total resistance "R_(P), is given by [Eq.(12.0.4)]`
`1/R_(p)=1/5+1/10+1/30=1/3`
Thus, `R_(P)=3Omega`