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

Explanation

Given, emf of the battery= 10 V(E), internal resistance= `3Omega(r)`
We know that the current in the circuit,
`I=E/(R+r)`
i.e., `0.5=10/(R+3)`
i.e., `R+3=10/(0.5)=2`
or `R=17Omega`
when the circuit is closed, the terminal voltage of the battery,
V =E-Ir
i.e., `V = 10-(0.5) (3)`
`V = 10 - 1.5 = 8.5 V`
or V = IR, i.e., V=0.5`xx` 17 = 8.5V