Answer –
We are given:
The EMF of the battery as E = 10 V
The internal resistance of the battery as R = 3 Ω
The current in the circuit as I = 0.5 A
Let the resistance of the resistor be R.
So, the current in the circuit can be found by using Ohm’s Law in the following manner –
\[I=\frac{E}{R+r}\]
Upon rearranging the above equation, we get –
\[R+r=\frac{E}{I}=\frac{10}{0.5}\]
\[R=20-3=17\Omega \]
Let the terminal voltage of the resistor be V.
Then, according to Ohm’s law we know that –
V = IR
Upon substituting values in the equation, we get
V = 0.5 × 17 = 8.5 V
Therefore, 17 Ω is the resistance of the resistor and the terminal voltage is 8.5 V.