Question 509: Current Electricity

Question

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of $5 \Omega$ the balancing length is 250 cm . When the cell is shunted by $20 \Omega$, the balancing length of potentiometer wire is 400 cm . The internal resistance , of the cell is

Accepted Answer

Let the e.m.f. of the cell be $E$ and internal resistance be $r$.
Step 1: Write the formula for balance condition
When cell is connected in parallel with a shunt $R$, the potential difference across it (for a balance length $l$ on potentiometer) is:
$$ V = k l $$
where $k$ is the potential gradient.
Current through the external resistance $R$:
$$ I = \frac{E}{R + r} $$
Potential difference across $R$:
$$ V = I R = \frac{E R}{R + r} $$
At balance, this $V$ is equal to $k l$:
$$ k l = \frac{E R}{R + r} $$
Step 2: Write equations for both conditions
Case 1:
$R_1 = 5\,\Omega$, $l_1 = 250$ cm
$$ k \cdot 250 = \frac{E \cdot 5}{5 + r} 	ag{1} $$
Case 2:
$R_2 = 20\,\Omega$, $l_2 = 400$ cm
$$ k \cdot 400 = \frac{E \cdot 20}{20 + r} 	ag{2} $$
Step 3: Divide (1) by (2) to eliminate $k$ and $E$
$$ \frac{k \cdot 250}{k \cdot 400} = \frac{\frac{E \cdot 5}{5 + r}}{\frac{E \cdot 20}{20 + r}} $$
$$ \frac{250}{400} = \frac{5}{5 + r} 	imes \frac{20 + r}{20} $$
$$ \frac{5}{8} = \frac{5 (20 + r)}{20 (5 + r)} $$
Step 4: Simplify and solve for $r$
Multiply both sides by $20(5 + r)$:
$$ \frac{5}{8} 	imes 20(5 + r) = 5(20 + r) $$
$$ \frac{100}{8}(5 + r) = 100 + 5r $$
$$ 12.5(5 + r) = 100 + 5r $$
$$ 62.5 + 12.5r = 100 + 5r $$
Subtract $5r$ from both sides:
$$ 62.5 + 7.5r = 100 $$
Subtract $62.5$:
$$ 7.5r = 37.5 $$
$$ r = \frac{37.5}{7.5} = 5\,\Omega $$
Final Answer:
$\boxed{5\,\Omega}$
So, the correct answer is
Option C: $5\,\Omega$
.

Answer

$3 \Omega$

Answer

$4 \Omega$

Answer

$5 \Omega$

Answer

$6 \Omega$

Current Electricity

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of 5 \Omega the balancing length is 250 cm . When the cell is shunted by 20 \Omega, the balancing length of potentiometer wire is 400 cm . The internal resistance , of the cell is

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