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

Explanation

Emf of the battery, `E = 10 V`
Internal resistance of the battery, `r = 3 Omega`
Current in the circuit, `I = 0.5 A`
Resistance of the resistor `= R`
The relation for current using Ohm's law is,
`I = (E)/(R+r)`
`R + r=(R)/(I)`
`=(10)/(0.5) = 20 Omega`
`:. R = 20 - 3 = 17 Omega`
Terminal voltage of the resistor = V
According to Ohm's law,
`V = IR`
`= 0.5 xx 17`
`= 8.5 V`
Therefore, the resistance of rthe resistor is `17 Omega` and the terminal voltage is 8.5 V.