Question 508: Current Electricity

Question

A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by $5 \Omega$. When a resistance of $15 \Omega$ is used for shunting, null point moves to 300 cm . The internal resistance of the cell is

Accepted Answer

Length of potentiometer wire with $5\,\Omega$ shunt, $l_1 = 200$ cm
Length of potentiometer wire with $15\,\Omega$ shunt, $l_2 = 300$ cm
Let the emf of the cell be $E$ and internal resistance $r$.
Let $R_1 = 5\,\Omega$ and $R_2 = 15\,\Omega$.
Concept:
At null point, potential difference across $R$ (shunt) is balanced by potential across $l$ cm of the potentiometer wire.
So,
$ \frac{E 	imes R_1}{R_1 + r} \propto l_1 $
$ \frac{E 	imes R_2}{R_2 + r} \propto l_2 $
Taking ratio:
$ \frac{\left(\frac{E R_1}{R_1 + r}ight)}{\left(\frac{E R_2}{R_2 + r}ight)} = \frac{l_1}{l_2} $
$ \frac{R_1}{R_1 + r} \cdot \frac{R_2 + r}{R_2} = \frac{l_1}{l_2} $
Now, cross-multiplied:
$ \frac{R_1 (R_2 + r)}{R_2 (R_1 + r)} = \frac{l_1}{l_2} $
Substitute the values:
$ \frac{5 (15 + r)}{15 (5 + r)} = \frac{200}{300} = \frac{2}{3} $
Now, solve for $r$:
Cross-multiplied:
$ 3 	imes 5 (15 + r) = 2 	imes 15 (5 + r) $
$ 15 (15 + r) = 10 (5 + r) $
$ 225 + 15 r = 50 + 10 r $
$ 15 r - 10 r = 50 - 225 $
$ 5 r = -175 $
$ r = -35 $
Check calculation:
Let's try again, carefully:
$ 3 	imes 5 (15 + r) = 2 	imes 15 (5 + r) $
$ 15 (15 + r) = 30 (5 + r) $
$ 225 + 15 r = 150 + 30 r $
$ 225 - 150 = 30r - 15r $
$ 75 = 15r $
$ r = \frac{75}{15} = 5 \Omega $
Final Answer:
The internal resistance of the cell is $\boxed{5\,\Omega}$.
Correct Option: Option C

Answer

$3 \Omega$

Answer

$4 \Omega$

Answer

$5 \Omega$

Answer

$6 \Omega$

Current Electricity

A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by 5 \Omega. When a resistance of 15 \Omega is used for shunting, null point moves to 300 cm . The internal resistance of the cell is

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