Two batteries, each of internal resistance \( 5\,\Omega \) and emf 10 V, are connected in parallel. Find the equivalent internal resistance and the equivalent emf of the combination.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \frac{5}{2}\,\Omega \), 10 V
Like resistors, cells joined in parallel can be replaced by one equivalent cell. For two cells of emfs \( \varepsilon_1, \varepsilon_2 \) and internal resistances \( r_1, r_2 \) in parallel:
\( \displaystyle r_{\mathrm{eq}}=\frac{r_1r_2}{r_1+r_2} \)
\( \displaystyle \varepsilon_{\mathrm{eq}}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2} \)
Here the two batteries are identical, each with:
\( r=5\,\Omega \)
\( \varepsilon=10\,\mathrm{V} \)
The equivalent internal resistance is:
\( \begin{aligned} r_{\mathrm{eq}} &= \frac{5\times5}{5+5}\,\Omega \\ &= \frac{5}{2}\,\Omega \end{aligned} \)
The equivalent emf is:
\( \begin{aligned} \varepsilon_{\mathrm{eq}} &= \frac{10\times5+10\times5}{5+5}\,\mathrm{V} \\ &= 10\,\mathrm{V} \end{aligned} \)
Therefore, two identical cells in parallel give the same emf as one cell, with half the internal resistance.