Question

In the circuit diagram given below, AB is a uniform wire of resistance 15 Ω and length 1 m. It is connected to a cell E₁ of emf 2V and negligible internal resistance and a resistance R. The balance point with another cell E₂ of emf 75 mV is found at 30 cm from end A. Calculate the value of R.

Answer 1

As per the figure,
Total current through the wire AB is given by

\[I = \frac{E}{R + r}\]

\[ = \frac{2}{R + 15}\]

The potential gradient of the wire is given by

\[= I \times \frac{15}{100}\]

\[ = \frac{2}{R + 15} \times \frac{15}{100}\]

As,  the balance point with cell E₂ of emf 75 mV is found at 30 cm from end A

\[\frac{2}{R + 15} \times 0 . 15 \times 30 = 75 \times {10}^{- 3} \]

\[\left( \frac{2}{75 \times {10}^{- 3}} \times 0 . 15 \times 30 \right) - 15 = R\]

\[R = 105 \Omega\]