A battery of emf 10 V and internal resistance `2 Omega` is connected to parallel combination of two identical resistors of resistance R. Current in the circuit is 0.5 A. What is the resistance of each resistor ? Find the terminal voltage of the battery.

Explanation

Here `R_1 = R_2 =R`
E = 10 V
`r = 3Omega`
`I = 0.5 A`
Total resistance of the circuit `R_T=E/I`
`R_(T) = 10/(0.5) = 20 Omega`
Let `R_P` is the resultant of `R_1 and R_2`
`1/(R_P)=1/R_1 +1/R_2 =1/R +1/R=2/R`
`R_P=R/2`
Now `R_P and r` are in series
So `R_("eff") = R_p +r (R)/2 +3`
But `R_("eff")= R_T`
`implies 20 = R/2 +3`
`R/2 =20-=17`
`R=34Omega`
Terminal potential difference, `V = E - Ir`
`V = 10 - 0.5(3)`
`V = 10-1.5`
V = 8.5 volt