Question:medium

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

Updated On: Apr 30, 2026
  • 1A
  • \(\frac{2}{3}A\)
  • \(\frac{3}{4}A\)
  • \(\frac{5}{6}A\)
Show Solution

The Correct Option is D

Solution and Explanation

To determine the effective electromotive force (emf) and internal resistance when two cells are connected in parallel, consider cells with \(E_1 = 2V\), \(r_1 = 1\Omega\) and \(E_2 = 1V\), \(r_2 = 2\Omega\). The formula for the effective emf \(E\) of cells in parallel is \(E = \frac{E_1r_2 + E_2r_1}{r_1 + r_2}\). Substituting the values yields \(E = \frac{2 \times 2 + 1 \times 1}{1 + 2} = \frac{4 + 1}{3} = \frac{5}{3}V\).

The effective internal resistance \(r\) is calculated using the formula \(r = \frac{r_1 \cdot r_2}{r_1 + r_2}\). With the given values, \(r = \frac{1 \times 2}{1 + 2} = \frac{2}{3}\Omega\).

The total resistance \(R_T\) is the sum of the effective internal resistance and the external resistance: \(R_T = r + R_{\text{ext}} = \frac{2}{3} + \frac{4}{3} = 2\Omega\).

Ohm's Law provides the current \(I\) through the circuit: \(I = \frac{E}{R_T}\). Therefore, \(I = \frac{\frac{5}{3}}{2} = \frac{5}{6}A\).

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

Was this answer helpful?
9


Questions Asked in CUET (UG) exam