Question

The current through a \(\frac{4}{3}\Omega\) external resistance connected to a parallel combinationof two cells of 2 V and 1 V emf and internal resistances of 1 \(\Omega\) and 2 \(\Omega\) respectively is ______.
Fill in the blank with the correct answer from the options given below

• A. 1A
• B. \(\frac{2}{3}A\)
• C. \(\frac{3}{4}A\)
• D. \(\frac{5}{6}A\)

Answer 1

The Correct Option is D

When two cells are connected in parallel, the effective emf and internal resistance of the combination need to be calculated. Let \(E_1 = 2V\) and \(r_1 = 1\Omega\), and \(E_2 = 1V\) and \(r_2 = 2\Omega\). The effective emf \(E\) of cells in parallel is given by: \(E = \frac{E_1r_2 + E_2r_1}{r_1 + r_2}\). 

Substituting the given values, \(E = \frac{2 \times 2 + 1 \times 1}{1 + 2} = \frac{4 + 1}{3} = \frac{5}{3}V\).

The effective internal resistance \(r\) is given by: \(r = \frac{r_1 \cdot r_2}{r_1 + r_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3}\Omega\).

Now we calculate the total resistance \(R_T\): \(R_T = r + R_{\text{ext}} = \frac{2}{3} + \frac{4}{3} = 2\Omega\).

The current \(I\) through the circuit is given by Ohm's Law: \(I = \frac{E}{R_T} = \frac{\frac{5}{3}}{2} = \frac{5}{6}A\).

Thus, the current through the external resistance is \(\frac{5}{6}A\).