Question 152: Current Electricity

Question

To measure the internal resistance of a battery, potentiometer is used. For $$R=10 \Omega$$, the balance point is observed at $$l=500 \mathrm{~cm}$$ and for $$\mathrm{R}=1 \Omega$$ the balance point is observed at $$l=400 \mathrm{~cm}$$. The internal resistance of the battery is approximately :

Accepted Answer

To measure the internal resistance of a battery using a potentiometer, we need to understand the principle behind it. The potentiometer is used to measure the voltage across the battery (emf) under different conditions. The balance length corresponds to the emf of the battery when no current is drawn, whereas under load, it corresponds to the terminal voltage.
Let's denote the emf of the battery by $$E$$ and the internal resistance by $$r$$. According to Ohm's law, when a resistance $$R$$ is connected across the battery, the terminal voltage $$V$$ is given by:
$$ V = E - Ir $$
where $$I$$ is the current through the circuit.
From the problem, the balance length $$l$$ is proportional to the voltage across the potentiometer wire. Thus, we can write:
$$ \frac{V_1}{V_2} = \frac{l_1}{l_2} $$
In the first condition, when $$R = 10 \Omega$$ and the balance length $$l_1 = 500 \, 	ext{cm}$$:
$$ V_1 = E - I_1 r $$
For the second condition, when $$R = 1 \Omega$$ and the balance length $$l_2 = 400 \, 	ext{cm}$$:
$$ V_2 = E - I_2 r $$
Given that:
$$ \frac{V_1}{V_2} = \frac{500}{400} = \frac{5}{4} $$
Now let's denote the internal emf of the battery as $$E$$. We can use OhmÃ¢ÂÂs Law in the calculation of $$I_1$$ and $$I_2$$:
$$ I_1 = \frac{E}{R_1 + r} = \frac{E}{10 + r} $$
$$ I_2 = \frac{E}{R_2 + r} = \frac{E}{1 + r} $$
Now, rewriting the values of $$V_1$$ and $$V_2$$ we have:
$$ V_1 = E - I_1 r = E \left(1 - \frac{r}{10 + r}ight) = E \frac{10}{10 + r} $$
$$ V_2 = E - I_2 r = E \left(1 - \frac{r}{1 + r}ight) = E \frac{1}{1 + r} $$
Using the ratio:
$$ \frac{V_1}{V_2} = \frac{\frac{10E}{10 + r}}{\frac{E}{1 + r}} = \frac{10 \cdot (1 + r)}{1 \cdot (10 + r)} = \frac{10 + 10r}{10 + r} $$
Given:
$$ \frac{V_1}{V_2} = \frac{5}{4} $$
So, we can set up the equation:
$$ \frac{10 + 10r}{10 + r} = \frac{5}{4} $$
Cross multiplying gives:
$$ 4(10 + 10r) = 5(10 + r) $$
Expanding both sides:
$$ 40 + 40r = 50 + 5r $$
Rearranging terms to solve for $$r$$:
$$ 40r - 5r = 50 - 40 $$
$$ 35r = 10 $$
$$ r = \frac{10}{35} = \frac{2}{7} \approx 0.285 \, \Omega $$
Since this value is closest to $$0.3 \Omega$$, the correct option is:
Option B: $$0.3 \Omega$$

Answer

$$0.1 \Omega$$

Answer

$$0.3 \Omega$$

Answer

$$0.2 \Omega$$

Answer

$$0.4 \Omega$$

Current Electricity

To measure the internal resistance of a battery, potentiometer is used. For $R=10 \Omega, the balance point is observed at l=500 \mathrm{~cm} and for \mathrm{R}=1 \Omega the balance point is observed at l=400 \mathrm{~cm}$. The internal resistance of the battery is approximately :

Choose the correct answer: