A battery of emf 10 v and internal resistane `3 Omega` is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistane of the resistors ? What is the terminal voltage of the battery when the circuit is closed ?

Explanation

Let R be the resistance of the resistor. Then the current in the circuit is given by: `I= (E )/(r+R)`
`rArr R= (E )/(I) =r`
Here `E= 10V`
`r=3 Omega, I= 0.5A`
`:. R= (10)/(0.5) -3= 17Omega`
Also, the terminal voltage of the battery when the circuit is closed is given by: `V= E-Ir`
Using values given above, we get
`V= 10-0.5 xx 3`
`rArr V= 8.5V`