Question 520: Current Electricity

Question

A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of $5 \Omega$. But when it is shunted by a resistance of $10 \Omega$, then balancing length increases by 25 cm . The balancing length when the cell is in an open circuit is

Accepted Answer

To solve for the balancing length when the cell is in an open circuit, we can use the equation related to the potentiometer setup:
$ r = s \left(\frac{l - l_b}{l_b}ight) $
Here, $ r $ refers to the internal resistance of the cell, $ s $ is the shunt resistance, $ l $ is the total balancing length in an open circuit, and $ l_b $ is the balancing length when the cell is shunted.
Given:
First condition: $ s = 5 \, \Omega $, $ l_b = 150 \, 	ext{cm} $
Second condition: $ s = 10 \, \Omega $, $ l_b = 175 \, 	ext{cm} $ (increased by 25 cm)
Equating the two scenarios:
$ 5 \left(\frac{l - 150}{150}ight) = 10 \left(\frac{l - 175}{175}ight) $
Simplifying the equation:
$ \frac{l - 150}{6} = 2 \left(\frac{l - 175}{7}ight) $
Multiply both sides to clear the fractions:
$ 7(l - 150) = 12(l - 175) $
Expanding both sides:
$ 7l - 1050 = 12l - 2100 $
Subtract $7l$ from both sides:
$ -1050 = 5l - 2100 $
Adding 2100 to both sides:
$ 1050 = 5l $
Solving for $ l $:
$ l = 210 \, 	ext{cm} $
Thus, the balancing length of the cell in an open circuit is $210 	ext{ cm} $.

Answer

200 cm

Answer

210 cm

Answer

225 cm

Answer

250 cm

Current Electricity

A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of 5 \Omega. But when it is shunted by a resistance of 10 \Omega, then balancing length increases by 25 cm . The balancing length when the cell is in an open circuit is

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