Two cells of emfs $1$ V and $2$ V and internal resistance $2\ \Omega$ and $1\ \Omega$, respectively connected in parallel, gave a current of $1$ A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\dfrac{\alpha}{5}$ A. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Concept. Cells in parallel act as one cell: $E_{\text{eq}}=\dfrac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}$, $r_{\text{eq}}=\dfrac{r_1r_2}{r_1+r_2}$. Reversing a cell changes the sign of its emf.
Step 1. Original. $E_{\text{eq}}=\dfrac{0.5+2}{1.5}=\dfrac53$ V, $r_{\text{eq}}=\dfrac23\ \Omega$; $1=\dfrac{5/3}{2/3+R}\Rightarrow R=1\ \Omega$.
Step 2. One reversed. $E_{\text{eq}}=\dfrac{-0.5+2}{1.5}=1$ V.
Step 3. Current. $\dfrac{1}{2/3+1}=\dfrac35$ A.
Answer. $\alpha=3$.
Solution by Sreeraj P, M.Sc Physics
