Question

Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.

Hint:

When cells are connected in series, their emfs add up and their internal resistances add up. When cells are connected in parallel, the equivalent emf and equivalent internal resistance are calculated using specific formulas. After finding the equivalent emf and equivalent internal resistance for both series and parallel configurations, use Ohm's law to find the total current in each case. Finally, calculate the ratio of the currents as required.

Answer 1

Correct Answer: 4

Case 1: Cells in series The equivalent emf of the cells in series is \( \varepsilon_{eq,s} = \varepsilon_1 + \varepsilon_2 = 1 \, \text{V} + 2 \, \text{V} = 3 \, \text{V} \). 
The equivalent internal resistance of the cells in series is \( r_{eq,s} = r_1 + r_2 = 2 \, \Omega + 1 \, \Omega = 3 \, \Omega \). 
The total resistance in the circuit is \( R_1 = r_{eq,s} + R = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega \). 
The total current in the circuit is \( I_1 = \frac{\varepsilon_{eq,s}}{R_1} = \frac{3 \, \text{V}}{9 \, \Omega} = \frac{1}{3} \, \text{A} \). 

Case 2: Cells in parallel The equivalent emf of the cells in parallel is \( \varepsilon_{eq,p} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{\frac{1+4}{2}}{\frac{1+2}{2}} = \frac{5/2}{3/2} = \frac{5}{3} \, \text{V} \). 
The equivalent internal resistance of the cells in parallel is \( r_{eq,p} = \frac{1}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{1}{\frac{1}{2} + \frac{1}{1}} = \frac{1}{\frac{1+2}{2}} = \frac{1}{3/2} = \frac{2}{3} \, \Omega \). 
The total resistance in the circuit is \( R_2 = r_{eq,p} + R = \frac{2}{3} \, \Omega + 6 \, \Omega = \frac{2 + 18}{3} \, \Omega = \frac{20}{3} \, \Omega \). 
The total current in the circuit is \( I_2 = \frac{\varepsilon_{eq,p}}{R_2} = \frac{5/3 \, \text{V}}{20/3 \, \Omega} = \frac{5}{20} \, \text{A} = \frac{1}{4} \, \text{A} \). 
Now we need to find the value of \( \frac{I_1}{I_2} \): \[ \frac{I_1}{I_2} = \frac{1/3}{1/4} = \frac{1}{3} \times \frac{4}{1} = \frac{4}{3} \] We are given that \( \frac{I_1}{I_2} = \frac{x}{3} \). Comparing the two expressions for \( \frac{I_1}{I_2} \): \[ \frac{x}{3} = \frac{4}{3} \] Therefore, the value of \( x \) is 4.