A cell of emf 2V and internal resistance `4 Omega` is connected across a parallel combination of two resistors of resistance `10 Omega` and `20 Omega`. Find the current through each resistor using Kirchhoff's laws.

Explanation

Data: `E=2V, r=4Omega, R_(1) = 10 Omega, R_(2)= 20Omega`
Let `I_(1)` and `I_(2)` be the current drawn from the cell as shown in the figure.
Then, at node B, by Kirchhoff's current law,
`I-I_(1-I_(2)=0` or `I=I_(1) + I_(2)`…………..(i)
Applying Kirchhoff's voltage law to loo ABCEFA, we get,
`-I_(1)R_(1) - Ir + E=0`
`therefore -10I_(1) - 4(I_(1)+I_(2))+2=0`........................ [From Eq.(1)]
`therefore 14I_(1) + 4I_(2)=2`
`therefore 7I_(1) + 2I_(2)=I`....................(2)

Applying Kirchoff's voltage law to loop BDECB, we get,
`-I_(2)R_(2) + I_(1)R_(1)=0`
`therefore I_(1) = 2I_(2)`..............(3)
Substitutiing for `I_(1)` in Eq.(2),
`(7 xx 2)I_(2) + 2I_(2)=1`
`therefore 14I_(2) + 2I_(2)=1`
`therefore 16I_(2)=1`
`therefore I_(2) = 1/16A= 0.0625 A`
Substituting for `I_(2)` in Eq.(3), we get,
`I_(1) = 2 xx 1/16 =1/8 A = 0.125A`
`therefore I= I_(1) + I_(2) = 0.125 + 0.0625 = 0.1875` A