The terminal voltage of the battery, whose emf is 10 V and internal resistance 1Ω, when connected through an external resistance of 4Ω as shown in the figure is:

Correct Answer: 8V

To find the terminal voltage of the battery, we analyze the circuit diagram provided in the image:

1. The dashed box represents the battery, which consists of an electromotive force (emf) source labeled as 10V and an internal resistance labeled as 1Ω.

2. The external resistor connected to this battery is labeled as 4Ω.

First, let us identify the given values:
- Electromotive force of the battery, $E=10\text{V}$
- Internal resistance of the battery, $r=1\Omega$
- External resistance, $R=4\Omega$

Step 1: Calculate the total current in the circuit
Since the internal resistance and the external resistor are in series, the total resistance of the circuit is:

$R_{\text{total}}=R+r=4\Omega +1\Omega =5\Omega$

Using Ohm's law, the current flowing through the circuit is:

$I=\frac{E}{R+r}=\frac{10\text{V}}{5\Omega }=2\text{A}$

Step 2: Calculate the terminal voltage of the battery
The terminal voltage ($V$) is the potential difference across the battery terminals when current is flowing. This can be calculated in two ways:

Method A: By calculating the potential drop across the external resistance:

$V=I\times R$

Substituting the values:

$V=2\text{A}\times 4\Omega =8\text{V}$

Method B: By subtracting the potential drop across the internal resistance from the emf of the battery:

$V=E-I\times r$

Substituting the values:

$V=10\text{V}-(2\text{A}\times 1\Omega )=10\text{V}-2\text{V}=8\text{V}$

Both methods yield the same result. Therefore, the terminal voltage of the battery is 8V.