Problem 15 - A Complete Resource Book in Physics for JEE Main

Question

A cell of emf \(E\) is connected across a resistance \(R\). The potential difference between the terminals of the cell is found to be \(V\). The internal resistance of the cell must be (A) \(\frac{2(E-V) V}{R}\) (B) \(\frac{2(E-V) R}{E}\) (C) \(\frac{(E-V) R}{V}\) (D) \((E-V) R\)

Step-by-Step Solution

Answer

The internal resistance of the cell is given by option (C) \(r = \frac{(E - V) R}{V}\).

Step 1: Use Kirchhoff's law and Ohm's law

: According to Kirchhoff's law, the sum of the emf and potential difference across the internal resistance of the cell is equal to the potential difference across the external resistance R. Let's denote the current passing through the circuit as I and the internal resistance of the cell as r. From Ohm's law, we know that \(I = \frac{V}{R}\), where V is the potential difference across the resistance R. Using Kirchhoff's law: \[E - Ir = V\]

Step 2: Substitute the value of current (I) from Ohm's law

: As \(I = \frac{V}{R}\), we can substitute it back into the equation from the first step: \(E - \frac{V}{R}r = V\) Now we will solve for r, the internal resistance of the cell.

Step 3: Rearrange the equation to solve for r

: We need to isolate r. First, multiply both sides by R to get rid of the fraction: \(ER - Vr = VR\) Now, we move all the terms containing r to one side and everything else to the other side: \(Vr = E - V)R\) Finally, we divide both sides by V to solve for r: \(r = \frac{(E - V)R}{V}\) Therefore, the internal resistance of the cell is given by the option (C) \(\frac{(E - V) R}{V}\).