Question

Two cells of emfs \(1\,\text{V}\) and \(2\,\text{V}\) and internal resistances \(2\,\Omega\) and \(1\,\Omega\) respectively connected in parallel, gave a current of \(1\,\text{A}\) through an external resistance. If the polarity of one cell is reversed, the current through the external resistance will be \( \frac{\alpha}{5} \) A. The value of \( \alpha \) is _____.

Answer 1

Correct Answer: 3

Concept: Equivalent emf of two cells in parallel: \[ E=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}} \] Equivalent internal resistance: \[ r=\frac{r_1r_2}{r_1+r_2} \] Step 1: {Find equivalent emf initially.} \[ E_1=1,\quad r_1=2 \] \[ E_2=2,\quad r_2=1 \] \[ E=\frac{\frac{1}{2}+2}{\frac{1}{2}+1} \] \[ =\frac{2.5}{1.5} \] \[ =\frac{5}{3}\,\text{V} \] Equivalent resistance: \[ r=\frac{2\times1}{2+1} \] \[ =\frac{2}{3}\,\Omega \] Step 2: {Use given current.} \[ I=1\,\text{A} \] \[ 1=\frac{E}{R+r} \] \[ R+r=\frac{5}{3} \] \[ R=\frac{5}{3}-\frac{2}{3}=1\,\Omega \] Step 3: {Reverse polarity of one cell.} Now \[ E_2=-2 \] Equivalent emf: \[ E'=\frac{\frac{1}{2}-2}{\frac{1}{2}+1} \] \[ =\frac{-1.5}{1.5} \] \[ =-1\,\text{V} \] Step 4: {Find new current.} \[ I=\frac{|E'|}{R+r} \] \[ =\frac{1}{1+\frac{2}{3}} \] \[ =\frac{1}{\frac{5}{3}} \] \[ =\frac{3}{5}\,\text{A} \] Thus \[ \frac{\alpha}{5}=\frac{3}{5} \] \[ \alpha=3 \]