Question

The terminal potential difference of a cell in the open circuit is 2 V. When the cell is connected to a 10$\omega$ resistor, the terminal potential difference falls to 1.5 V. The internal resistance of the cell is:

• A. \( \frac{10}{3} \, \Omega \)
• B. \( \frac{10}{9} \, \Omega \)
• C. \( \frac{20}{7} \, \Omega \)
• D. \( \frac{15}{6} \, \Omega \)
• E. \( \frac{13}{2} \, \Omega \)

Hint:

Use the relationship between terminal voltage, emf, current, and internal resistance to solve problems involving internal resistance of a cell.

Answer 1

The Correct Option is A

Step 1: The terminal potential difference \( V \) is related to the emf \( E \), the internal resistance \( r \), and the external resistance \( R \) by: \[ V = E - I r \] where \( I \) is the current. 
Step 2: The current \( I \) is: \[ I = \frac{E}{R + r} \] Substitute this into the equation for \( V \): \[ V = E - \frac{E r}{R + r} \] Step 3: Given \( E = 2 \, {V} \), \( V = 1.5 \, {V} \), and \( R = 10 \, \Omega \), we can solve for \( r \): \[ 1.5 = 2 - \frac{2r}{10 + r} \] Step 4: Solving for \( r \): \[ 0.5 = \frac{2r}{10 + r} \quad \Rightarrow \quad 0.5(10 + r) = 2r \quad \Rightarrow \quad 5 + 0.5r = 2r \] \[ 5 = 1.5r \quad \Rightarrow \quad r = \frac{10}{3} \, \Omega \] Thus, the internal resistance of the cell is \( \frac{10}{3} \, \Omega \).