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.

Answer 2

This problem involves calculating the current in a circuit for two different configurations of cells (series and parallel) and then finding the ratio of these currents to determine the value of a given variable.

Concept Used:

The key concepts are the calculation of equivalent EMF and equivalent internal resistance for series and parallel combinations of cells.

1. Series Combination:

For two cells with EMFs \(E_1, E_2\) and internal resistances \(r_1, r_2\) connected in series, the equivalent EMF (\(E_{eq}\)) and equivalent internal resistance (\(r_{eq}\)) are:

\[ E_{eq} = E_1 + E_2 \] \[ r_{eq} = r_1 + r_2 \]

The total current \(I\) in a circuit with an external resistance \(R\) is given by Ohm's law for a complete circuit:

\[ I = \frac{E_{eq}}{R + r_{eq}} \]

2. Parallel Combination:

For the same two cells connected in parallel, the equivalent EMF and internal resistance are:

\[ E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \] \[ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \]

The total current \(I\) is again given by \(I = \frac{E_{eq}}{R + r_{eq}}\).

Step-by-Step Solution:

Step 1: Calculate the current \(I_1\) for the series combination. First, find the equivalent EMF and internal resistance.

Given values: \(E_1 = 1 \text{ V}\), \(r_1 = 2 \, \Omega\), \(E_2 = 2 \text{ V}\), \(r_2 = 1 \, \Omega\), and \(R = 6 \, \Omega\).

\[ E_{eq, series} = E_1 + E_2 = 1 \text{ V} + 2 \text{ V} = 3 \text{ V} \] \[ r_{eq, series} = r_1 + r_2 = 2 \, \Omega + 1 \, \Omega = 3 \, \Omega \]

Now, calculate the current \(I_1\):

\[ I_1 = \frac{E_{eq, series}}{R + r_{eq, series}} = \frac{3 \text{ V}}{6 \, \Omega + 3 \, \Omega} = \frac{3}{9} \text{ A} = \frac{1}{3} \text{ A} \]

Step 2: Calculate the current \(I_2\) for the parallel combination. First, find the equivalent EMF and internal resistance.

\[ E_{eq, parallel} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \frac{(1)(1) + (2)(2)}{2 + 1} = \frac{1 + 4}{3} = \frac{5}{3} \text{ V} \] \[ r_{eq, parallel} = \frac{r_1 r_2}{r_1 + r_2} = \frac{(2)(1)}{2 + 1} = \frac{2}{3} \, \Omega \]

Now, calculate the current \(I_2\):

\[ I_2 = \frac{E_{eq, parallel}}{R + r_{eq, parallel}} = \frac{\frac{5}{3}}{6 + \frac{2}{3}} = \frac{\frac{5}{3}}{\frac{18+2}{3}} = \frac{\frac{5}{3}}{\frac{20}{3}} = \frac{5}{20} \text{ A} = \frac{1}{4} \text{ A} \]

Step 3: Calculate the ratio \(\frac{I_1}{I_2}\).

\[ \frac{I_1}{I_2} = \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{1}{3} \times 4 = \frac{4}{3} \]

Step 4: Compare this ratio with the given expression \(\frac{x}{3}\) to find the value of x.

\[ \frac{I_1}{I_2} = \frac{x}{3} \] \[ \frac{4}{3} = \frac{x}{3} \]

By comparing both sides, we get:

\[ x = 4 \]

The value of x is 4.