Question

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

Answer 1

Hint: Draw the circuit diagram according to the question. Internal resistance can be considered to be a series resistance with the battery. Use Kirchhoff's Voltage Law to write the voltage equation for the circuit. Solve it to find the resistor. Terminal voltage of the battery is given by the multiplication of load current and load resistance.

Formula Used:
Ohm’s law gives us,
V=IR
Where,
V is the voltage across the resistor
I is the current through the resistor
R is the resistance of the resistor

Complete step by step answer:
Let us first, look at the circuit diagram.

Let’s assume that the current flowing through the circuit is i.
So, we can write,

$\Rightarrow R={\displaystyle \frac{E-Ir}{I}}$

The value of the external resistance is R

The EMF of the battery is 10 V

Hence, we can write,
$R={\displaystyle \frac{E-Ir}{I}}$

Internal resistance of the battery, r =3Ω.

We can write the following equation using Kirchhoff’s voltage law,
$E-IR-Ir=0$
$\Rightarrow R={\displaystyle \frac{10-(0.5)3}{0.5}}=17$

Hence, the value of Resistance R = 17 Ω

Now, we can find the voltage across the battery by calculating the voltage across the load resistor.
Voltage across the battery and voltage across the load resistor should be the same because we can consider them to be in parallel connection.

Hence, we can write,
The terminal voltage of the battery = The load voltage =
$V=IR=(0.5)(17)=8.5$

So, the terminal voltage of the battery is 8.5 V.

Note: The terminal voltage of the battery is less than the EMF of the battery because of the presence of internal resistance. We can find the terminal voltage in another method as well.

The terminal voltage will be given by,
$V=E-Ir=10-(0.5)3=8.5$

Some of the EMF is lost in the voltage drop across the internal resistance.