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Question: Answered & Verified by Expert
Two batteries of emf $\varepsilon_1$ and $\varepsilon_2\left(\varepsilon_2>\varepsilon_1\right)$ and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure.
PhysicsCurrent Electricity
Options:
  • A
    Two equivalent emf $\varepsilon_{\text {eq }}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$, i.e., $\varepsilon_1 < \varepsilon_{\mathrm{eq}} < \varepsilon_2$
  • B
    The equivalent emf $\varepsilon_{\mathrm{eq}}$ is smaller than $\varepsilon_1$
  • C
    The $\varepsilon_{\mathrm{eq}}$ is given by $\varepsilon_{\mathrm{eq}}=\varepsilon_1+\varepsilon_2$ always
  • D
    $\varepsilon_{e q}$ is independent of internal resistances $r_1$ and $r_2$
Solution:
2003 Upvotes Verified Answer
The correct answer is:
Two equivalent emf $\varepsilon_{\text {eq }}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$, i.e., $\varepsilon_1 < \varepsilon_{\mathrm{eq}} < \varepsilon_2$
As we know the equivalent $\operatorname{emf}\left(\varepsilon_{\mathrm{eq}}\right)$ in the parallel combination
$$
\varepsilon_{\text {eq }}=\frac{\varepsilon_2 r_1+\varepsilon_1 r_2}{r_1+r_2}
$$
So according to formula the equivalent emf $\varepsilon_{\mathrm{eq}}$ of the two cells in parallel combination is between $\varepsilon_1$ and $\varepsilon_2$. Thus $\left(\varepsilon_1\right.$ $ < \varepsilon_{\text {eq }} < \varepsilon_2$ ).

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