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Grouping Of Cells - Practice Questions & MCQ

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

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

  • 50 Questions around this concept.

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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

The current flowing through $3.25 \Omega$ resistance is 

5 cells of emf E and internal resistance r are connected in parallel giving 1A current. Then current from 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:

A house is served by a 220 V supply line in a circuit protected by a 9 ampere fuse. The maximum number of 60 W lamps in parallel that can be turned on, 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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