Tag: combinations of components

Questions Related to combinations of components

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. $1.22\ V$

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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

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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. $r _1+r _2$

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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

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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. $1.5\Omega$

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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

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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. $n/r$

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

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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.