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Combination of cells - class-XI

Description: combination of cells
Number of Questions: 20
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Tags: physics electricity and magnetism current electricity electric current through conductors electric circuits
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For a resistance R and capacitance C in series the impedence is twice that of a parallel combination of the same elements.The frequency of the applied emf shall be.

physics electric current through conductors combination of resistors combination of cells combinations of components

  1. $\dfrac{{2\pi }}{{Rc}}$

  2. $\dfrac{1}{{2\pi {R _1}c}}$

  3. $\dfrac{{2\pi }}{{\sqrt {RC} }}$

  4. $\dfrac{1}{{2\pi \sqrt {RC} }}$

Show answer
Correct Option: A

Two non ideal batteries are connected in series. Consider the following statements
(A) The equivalent emf is larger than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances.

physics electric current through conductors combination of resistors combination of cells combinations of components

  1. Both A and B are correct

  2. A is correct but B is wrong

  3. B is correct but A is wrong

  4. Both A and B wrong

Show answer
Correct Option: B
Explanation:

If we add two non ideal batteries in series then
so in case (a) the equivalent e.m.f.  may be larger than either of two e.m.f., but in case (b) the equivalent e.m.f. may be smaller than either of two e.m.f.
In series total resistance is always greater than individual resistance, whether batteries are connected in any way 

[i.e. either according to case (a) or case (b)].

When cells are connected in series greater emf is obtained in the circuit.

physics electric current through conductors combination of resistors combination of cells combinations of components

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

When two or more cells are connected in series in a circuit, the total potential difference is the sum of their potential differences. Hence, the cells are connected in series in order to produce a larger total emf.

Two cells, having the same e.m.f are connected in series through an external resistance R. Cells have internal resistance $r _1$ and $r _2$ ( $r _1$ > $r _2$) respectively. When the circuit is closed, the potential difference across the first cell is zero. the value of R is

physics electric current through conductors combination of resistors combination of cells combinations of components

  1. $\dfrac{r _1 + r _2}{2}$

  2. $\dfrac{r _1 - r _2}{2}$

  3. ${r _1 + r _2}$

  4. ${r _1 - r _2}$

Show answer
Correct Option: D
Explanation:
Two cells of emf E are connected in series through external resistance R.
Thus, total resistance is R + r1 + r2
 
Thus, current across the circuit is total voltage divided by total resistance.
I=E+E/R+r1+r2
I=2E/R+r1+r2
 
Now, when circuit is closed, the voltage across first cell is zero. Thus, we have
Ir1=E
E/r1=2E/R+r1+r2
R=r1-r2
This is the value of R in terms of r1 and r2.

Which of the following statement(s) is/are correct?

physics electric current through conductors combination of resistors combination of cells combinations of components

  1. If n identical cells are connected in series and then the battery thus formed is short circuited by a conduction wire, current through the wire will be independent of n.

  2. If n identical cells are connected in parallel and then the battery thus formed is short circuited by a conducting wire, current through the wire will be directly proportional to n.

  3. If n identical cells are connected in parallel and then the battery thus formed is short circuited by a wire having a constant resistance, current through the wire will be directly increase as n increases.

  4. None of the above.

Show answer
Correct Option: A,B
Explanation:

In series, $I=\frac{nE}{nr}=\frac{E}{r}$ (i.e., I is 
independent of n)

In parallel, $I=\frac{E}{r/n}=\frac{nE}{r}, i.e,.\frac{nE}{r}, 
i.e.I\propto n$

After shorting by a wire having a constant resistance 
$I=E\left (R+\frac{r}{n}  \right )$

A battery of 20 cells (each having e.m.f. 1.8 volt and internal resistance 0.1 ohm) is charged by 220 volts and the charging current is 15A. The resistance to be put in the circuit is

physics electric current through conductors combination of resistors combination of cells combinations of components

  1. 10.27 ohms

  2. 12.27 ohms

  3. 8.62 ohms

  4. 16.24 ohms

Show answer
Correct Option: A
Explanation:

The total emf of the 20 cells connected in series are $1.8\times 20=36 V$. 

Hence, the total voltage in the circuit is $220V - 36V  =  184 V.$
From the Ohm's law, $V=IR$. Substituting the values of Voltage and current in the equation

 $R=\dfrac { V }{ I } ,\quad we\quad get\quad R=\dfrac { 184 }{ 15 } =12.27\quad ohms$.

The internal resistance of the 20 cells is given as  $20\times 0.1ohms=2 ohms$.
So, the total resistance R in the circuit is $ =12.27-2 ohms = 10.27 ohms.$
Hence, the resistance in the circuit is 10.27 ohms.

A simple electric motor has an armature resistance of $1\Omega$ and runs from a d.c. source of $12$V. It draws a current of $2$A when unloaded. When a certain load is connected to it, its speed reduces by $10\%$ of its initial value. The current drawn by the loaded motor is?

physics electric circuits combination of resistors combination of cells combinations of components

  1. $3$A

  2. $6$A

  3. $2$A

  4. $1$A

Show answer
Correct Option: A
Explanation:

The back emf$= V-IR$ 
$= 12-2 = 10V$
$10%$ of this emf is $1V$
Hence the current is $IR = 1\times1= 1A$
Hence the current drawn by the loaded motor is $1+2=3A$

When two identical cell are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is:

physics electric circuits combination of resistors combination of cells combinations of components

  1. 1.2

  2. 2

  3. 4

  4. 4.8

Show answer
Correct Option: C

A series battery of six lead accumulators, each of emf 2.0V and internal resistance 0.50$\Omega $ is charged by a 100 V dc supply. The series resistance should be used in the charging circuit in order to limit the current to 8.0A is

physics electric circuits combination of resistors combination of cells combinations of components

  1. 4$\Omega $

  2. 6$\Omega $

  3. 8$\Omega $

  4. 10$\Omega $

Show answer
Correct Option: C
Explanation:

$ \Rightarrow  (100-12) = I[3+R]$

$ \dfrac{88}{3+R} = 8 $

$ \Rightarrow  R = 8\Omega$

$n$ cells of each of EMF $E$ and internal resistance $r$ send the same current $R$ whether the cells are connected in series or parallel, then :

physics electric circuits combination of resistors combination of cells combinations of components

  1. $R = nr$

  2. $R = r$

  3. $r = nR$

  4. $ R = \sqrt {n}  \dfrac{E}{r}$

Show answer
Correct Option: D

A thin uniform wire $50\ cm$ long and of $1\ ohm$ resistance is connected to the terminals of an accumulator of $emf\ 2.2\ volt$ and the internal resistance $0.1\ ohm$. If the terminal of another cell can be connected to two point $26\ cm$ apart on the wire without altering the current in the wire, the emf of the cell is

physics electric circuits combination of resistors combination of cells combinations of components

  1. $1.12\ V$

  2. $1.04\ V$

  3. $1.18\ V$

  4. $1.22\ V$

Show answer
Correct Option: D

Two non-ideal batteries are connected in series. Consider the following statements:
(A) The equivalent emf is larger than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances

physics electric circuits combination of resistors combination of cells combinations of components

  1. Each of A and B is correct

  2. A is correct but B is wrong

  3. B is correct but A is wrong

  4. Each of A and B is wrong

Show answer
Correct Option: B
Explanation:

When two non-ideal batteries are connected in series, their equivalent emf is
$E _{eq} = E _1 + E _2$
Hence, the equivalent emf is larger than either of the two emfs.
Also, their equivalent internal resistance is
$r _{eq} = r _1 + r _2$
Hence, the equivalent resistance is larger than either of the two internal resistances.
Therefore, statement A is correct but B is wrong.

Two cells of same emf are connected in series. Their internal resistances are $r _1$ and $r _2$ respectively and $r _1 > r _2$. When this combination is connected to an external resistance R then the potential difference between the terminals of first cell becomes zero. In this condition the value of R will bw

physics electric circuits combination of resistors combination of cells combinations of components

  1. $\frac {r _1-r _2}{2}$

  2. $\frac {r _1+r _2}{2}$

  3. $r _1-r _2$

  4. $r _1+r _2$

Show answer
Correct Option: C
Explanation:

Let emf of each cell is $E$.
As they are connected in series so current in circuit is $I=\dfrac{E+E}{r _1+r _2+R}=\dfrac{2E}{r _1+r _2+R}$ 

Potential across terminal of first cell is $V _1=E-Ir _1=E-\dfrac{2Er _1}{r _1+r _2+R}$ 

As $V _1=0 \Rightarrow E-\dfrac{2Er _1}{r _1+r _2+R}=0$ 

$r _1+r _2+R-2r _1=0$ 

$R=r _1-r _2$

$24$ cells, each having the same e.m.f. and $2$ ohm internal resistance, are used to draw maximum current through an external resistance of $3$ ohm. The cells should be connected :

physics electric circuits combination of resistors combination of cells combinations of components

  1. In series

  2. in parallel

  3. In $4$ rows, each row having $6$ cells

  4. In $6$ rows, each row having $4$ cells

Show answer
Correct Option: A
Explanation:

(A) Option in series 

Net EMF = $n\varepsilon$
= 24$\epsilon$
Net resistance =nr
= 24$\times$2
=48$\Omega$
Current = $\dfrac { 24\varepsilon  }{ nr+R } $
$ =\dfrac { 24\varepsilon  }{ nr+R } $
$=\dfrac { 24\varepsilon  }{ 48+3 } $
 $=\dfrac { 24\varepsilon  }{ 51 } =0.47\varepsilon $
(B) Net EMF = $\varepsilon$ = (In parallel EMF is same )
Net resistance $=\dfrac { r }{ n } =\dfrac { 1\Omega  }{ 12 } $
Current $=\dfrac { \varepsilon  }{ \dfrac { 1 }{ 12 } +3 } $
$=\dfrac { 12\varepsilon  }{ 37 } =0.32\varepsilon $
(C) Option
net EMF in each row$ = n\varepsilon$
$= 6\varepsilon$
net resistance in each row $=n\varepsilon$
$= 12\Omega$
net EMF will be $6\varepsilon$ ( this will be in parallel)
Net resistance $\dfrac{r}{n}$
$=\dfrac{12}{4}=3\Omega$
Net current $=\dfrac{6\varepsilon}{3+3}$
$= 1\varepsilon A$
(D) Option 
Net EMF in each row $n\varepsilon$
=$4\varepsilon$
Net resistance in each row = nr
= 4$\times$2
= 8$\Omega$
Net EMF will be $4\varepsilon$ ( this will be in parallel)
Net resistance $\dfrac{r}{n}=\dfrac{8}{6}=\dfrac{4}{3}\Omega$
Current $ \dfrac { 4\varepsilon  }{ \dfrac { 4 }{ 3 } +3 } =\dfrac { 12\varepsilon  }{ 13 } =0.92\varepsilon $
Hence current will be maximum in (C) option

n identical cells are joined in series with its two cells A and B in the loop with reversed polarities.
EMF of each shell is E and internal resistance r. Potential difference across cell A or B is:

physics electric circuits combination of resistors combination of cells combinations of components

  1. $\dfrac{2E}{n}$

  2. $2E|1-\dfrac{1}{n}|$

  3. $\dfrac{4E}{n}$

  4. $2E|1-\dfrac{2}{n}|$

Show answer
Correct Option: C
Explanation:

Number of cells $=n$

EMF of each cell $=E$
Internal resistance of each cell, $= r$

All the cells are in series.
Thus,
Total internal resistance, $R=nr$
Total EMF, $e=nE$

As the two cells are connected in reverse polarity, these two cells will cancel-out the contribution of other two cells in the loop.

Then, the net EMF across the circuit will be
$E _{eq}=nE-4E$

The current $I$ in the circuit $=\dfrac{nE-4E}{nr}$

Voltage across the opposite connected batteries, $V=E-Ir$
$V=E-\dfrac{E(n-4)}{nr}\times r$
$V=\dfrac{4E}{n}$

Therefore, the voltage across the battery A or B will be $\dfrac{4E}{n}$.


Two similar cells, whether joined in series or in parallel, have the same current through an external resistance of $2\Omega$. The internal resistance of each cell is

physics electric circuits combination of resistors combination of cells combinations of components

  1. $1\Omega$

  2. $2\Omega$

  3. $0.5\Omega$

  4. $1.5\Omega$

Show answer
Correct Option: B
Explanation:

In series, current, $i _{1} = \dfrac {2E}{2 + 2r}$
In parallel, current, $i _{2} = \dfrac {E}{2 + \dfrac {r}{2}} = \dfrac {2E}{4 + r}$
According to the question
Since, $i _{1} - i _{2} \Rightarrow \dfrac {2E}{4 + r} = \dfrac {E}{2 + 2r}$
$\Rightarrow r = 2\Omega$.

36 identical cell each having emf 1.5 volt and internal resistance $ 0.5 \Omega$ are connected in series with an external resistance of $12 \Omega$ .If 6 cells are wrongly connected then current through the circuit will be 

physics electric circuits combination of resistors combination of cells combinations of components

  1. 1.2 A

  2. 1 A

  3. 2 A

  4. 4 A

Show answer
Correct Option: A
Explanation:

Total resistance in the circuit is, $R _T=36\times0.5+12=30\Omega$

Net EMF in the circuit is, $E _T=(30-6)\times1.5=36V$

So, current in the circuit is, $I=\dfrac{E _T}{R _T}=\dfrac{36}{30}=1.2A$

When n number identical cell of emf E and internal resistance is connected in series, the net internal resistance of the system will be  

physics electric circuits combination of resistors combination of cells combinations of components

  1. $\dfrac{nEr}{1+E}$

  2. $n^2r$

  3. $nr$

  4. $n/r$

Show answer
Correct Option: C
Explanation:

It is a common fact that when current flows the circuit, connected with a magnetic compass, the magnetic needle in the compass will show deflection due to the magnetic effect of the current. Magnetic lines of forces are created around the coil of wire and this caused the needle to deflect according to the direction of the current.
In this case, when the number of cells in the circuit is increased, the current through the circuit is also increased. This is because, the increase in cells indicates more current in the circuit. 
Hence, as the current in the circuit is increased, the deflection in the magnetic compass also increase further.


A cell of constant emf first connect to a resistance $R _1$ and then to connected to the resistance $R _2.$ If power delivered in both cases in the same then internal resistance of the cell

physics electric circuits combination of resistors combination of cells combinations of components

  1. $\dfrac { R _ { 1 } - R _ { 2 } } { 2 }$

  2. $ { R _ { 1 } + R _ { 2 } } $

  3. $\sqrt { R _ { 1 } R _ { 2 } }$

  4. $\sqrt { R _ { 1 } + R _ { 2 } / 2 }$

Show answer
Correct Option: A
Explanation:

The cell has constant emf$,$

Resistance is $R _1$ and $R _2$

$P = V _2 / R _1 + r$

$P = V _2 / R _2 - r$

$( R _1- R _2 ) / 2$

The internal resistance of the cell is $( R _1 - R _2 ) / 2$

Hence,

option $(A)$ is correct answer.

A battery is delivering same power to resistance $R _1$ and $R _2$ . Then find the internal resistance of battery :

physics electric circuits combination of resistors combination of cells combinations of components

  1. $\dfrac { R _ { 1 } - R _ { 2 } } { 2 }$

  2. ${ R _ { 1 } + R _ { 2 } }$

  3. $ \sqrt { R _ { 1 } + R _ { 2 } }$

  4. $\sqrt { R _ { 1 } + R _ { 2 } / 2 }$

Show answer
Correct Option: A
Explanation:
$\begin{array}{l} P=\frac { { { V^{ 2 } } } }{ { { R^{ 2 } }+r } } -------\left( 1 \right)  \\ P=\frac { { { V^{ 2 } } } }{ { { R^{ 2 } }-r } } -------\left( 2 \right)  \\ solve\, \, eqation\left( 1 \right) and\left( 2 \right)  \\ =\dfrac { { { R _{ 1 } }-{ R _{ 2 } } } }{ 2 }  \end{array}$
Hence,
option $(A)$ is correct answer.
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