Question

The emf of a cell is $$\epsilon$$ and its internal resistance is r. Its terminals are connected to a resistance R. The potential difference between the terminals is 1.6 V for $$R=4\mathrm{\Omega }$$ and 1.8 V for $$R=9\mathrm{\Omega }$$. Then
A. $$\epsilon =1V,r=1\mathrm{\Omega }$$
B. $$\epsilon =2V,r=1\mathrm{\Omega }$$
C. $$\epsilon =2V,r=2\mathrm{\Omega }$$
D. $$\epsilon =2.5V,r=0.5\mathrm{\Omega }$$

Answer 1

Hint:You can use Kirchhoff’s law to determine the current flowing through the circuit. The potential difference across the terminals of the battery is equal to the potential difference across the resistance. Write the two equations using this fact and solve them simultaneously to determine the internal resistance of the battery.

Formula used:
Potential difference, $$V=IR$$,
V is the voltage, I is the current and R is the resistance.

Complete step by step answer:
We know the current flowing through the circuit containing a resistor R and a battery of internal resistance r is given as,
$$I={\displaystyle \frac{\epsilon }{R+r}}$$, where, $$\epsilon$$ is the emf of the battery.
We can see the current flowing through resistance $$R_1=4\mathrm{\Omega }$$ is $$I_1={\displaystyle \frac{\epsilon }{R_1+r}}$$and we know that the same current is provided by the battery. Therefore, we can write, the potential difference across the battery is equal to the potential difference across the resistance $$R_1$$.
$$I_1{R}_1=\left({\displaystyle \frac{\epsilon }{R_1+r}}\right)R_1$$
$$\Rightarrow 1.6=\left({\displaystyle \frac{\epsilon }{R_1+r}}\right)R_1$$
Substituting $$R_1=4\mathrm{\Omega }$$ in the above equation, we get,
$$1.6=\left({\displaystyle \frac{\epsilon }{4+r}}\right)\times 4$$
$$\Rightarrow 0.4=\left({\displaystyle \frac{\epsilon }{4+r}}\right)$$
$$\Rightarrow \epsilon =1.6+0.4r$$ …… (1)
Again, we have given that the potential difference between the terminals is 1.8 V when $$R_2=9\mathrm{\Omega }$$. Therefore, we can write,
$$1.8=\left({\displaystyle \frac{\epsilon }{9+r}}\right)\times 9$$
$$\Rightarrow 0.2=\left({\displaystyle \frac{\epsilon }{4+r}}\right)$$
$$\Rightarrow \epsilon =1.8+0.2r$$ …… (2)
Equating equation (1) and (2), we get,
$$1.6+0.4r=1.8+0.2r$$
$$\Rightarrow 0.2r=0.2$$
$$\Rightarrow r=1\mathrm{\Omega }$$
Substituting $$r=1\mathrm{\Omega }$$ in equation (1), we get,
$$\epsilon =1.6+0.4\left(1\right)$$
$$\therefore \epsilon =2\text{V}$$

So, the correct answer is option B.

Note:The potential difference across the battery terminals is the product of current supplied by the battery and resistance of the circuit and not the internal resistance of the battery. In many cases, the internal resistance of the battery is negligible compared to the external resistance of the circuit and therefore, we often neglect the internal resistance. But it should not be neglected if the external resistance of the battery is of few ohms.