Question

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

Answer 1

Correct Answer: 4 - 5

Step 1: From the first equation:

\(\frac{\epsilon}{r+5} \times 5 = 200x\)
\(\frac{\epsilon}{r+5} = \frac{200x}{5} = 40x\)
\(\epsilon = 40x(r+5)\)

Step 2: Now substitute \(\epsilon = 40x(r+5)\) into the second equation:

\(\frac{\epsilon \times 15}{r+15} = 300x\)
Substituting \(\epsilon = 40x(r+5)\) into the above equation:
\(\frac{40x(r+5) \times 15}{r+15} = 300x\)
Simplifying:
\(\frac{600x(r+5)}{r+15} = 300x\)

Step 3: Cancel out the common term of \(x\):

\(\frac{600(r+5)}{r+15} = 300\)
Simplify the equation:
\(600(r+5) = 300(r+15)\)

Step 4: Expand both sides:

\(600r + 3000 = 300r + 4500\)

Step 5: Solve for r:

\(600r - 300r = 4500 - 3000\)
\(300r = 1500\)
\(r = \frac{1500}{300} = 5 \, \Omega\)

Thus, the value of r is: 5 Ω