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    JEE Main 2026 April 6 Shift 1 Question Paper with Solutions PDF

    Grouping Of Cells - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • Series and Parallel Grouping of cell is considered one of the most asked concept.

    • 27 Questions around this concept.

    Solve by difficulty

    The figure shows three circuits I, II and III which are connected to a 3V battery.  If the powers dissipated by the configurations I, II and III are P1, P2 and P3 respectively, then :

     

    The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2\Omega$. The value of the internal resistance of each cell is

    If n cells are connected in series and such m connections are made in parallel then the potential difference across each cell is:

    Concepts Covered - 1

    Series and Parallel Grouping of cell

     series grouping of cell:

    In series grouping anode of one cell is connected to the cathode of other cells

    $n=$ identical cells which are connected in series, then
    - Equivalent e.m.f of combination is $E_{e q}=n E$
    - Equivalent internal resistance $r_{e q}=n r$
    - Main current / current from each cell $i=\frac{n E}{R+n r}$
    - Power dissipated in the external circuit is $\left(\frac{n E}{R+n r}\right)^2 \cdot R$
    - Conditions for Maximum Power is $R=n r$
    - $P_{\text {max }}=n\left(\frac{E^2}{4 r}\right)$ when $n r<R$

     Parallel grouping of cell

    In parallel grouping, all anodes are connected to one point and all cathode together at other points

    For n cells connected in parallel
    Equivalent e.m.f $E_{e q}=E$
    Equivalent internal resistance $R_{e q}=\frac{r}{n}$
    The main current is

    $
    i=\frac{E}{R+\frac{r}{n}}
    $
    The potential difference across the external resistance

    $
    V=i R
    $
    Current from each cell

    $
    i^{\prime}=\frac{i}{n}
    $


    The power dissipated in the circuit

    $
    P=\left(\frac{E}{R+\frac{r}{n}}\right)^2 \cdot R
    $

    Condition for Maximum Power

    $\begin{aligned} & R=\frac{r}{n} \\ & P_{\max }=n\left(\frac{E^2}{4 r}\right)_{\text {when } r>>n R}\end{aligned}$

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    Series and Parallel Grouping of cell

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