A cell has an emf of 1.5V and an internal resistance 1`Omega`. It is connected to two resistances 2`Omega` and 3`Omega` in series. Find (a) the current flowing in the circuit, and (b) potential difference across the ends of each resistance.

Explanation

Data supplied `R_(1)=Omega," " R_(2)=3Omega, " " r=1Omega`
Total resistance in the circuite=1+2+3=6`Omega`
(a) `I=(V)/(R)=(1.5)/(6)=0.25A`
P.D across `2Omega=Rxx1=2xx0.25=0.5V`
P.D across `3Omega=Rxx1=3xx0.25=0.75V`
P.D across `1Omega=Rxx1=1xx.25=0.25V`