Tag: physics

Questions Related to physics

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.

State whether given statement is True or False
An electric bulb is a source of electricity.

physics simple electric circuits bulbs bulb bulb connected to an electric cell

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Electric Bulb is a source of light. Hence above statement is False.

Fill in the blanks:

Pure tungsten has _________.

physics simple electric circuits bulbs bulb bulb connected to an electric cell

  1. low resistivity and a high melting point.

  2. high resistivity and a high melting point.

  3. high resistivity and a low melting point.

  4. low resistivity and a low melting point.

Show answer
Correct Option: B
Explanation:

tungsten has a high melting point and is able to withstand heat and corrosion. Tungsten is one of the most durable types of filaments used in light bulbs. It must be exposed to very high temperatures before melting, and only carbon arc has a higher heat resistance level than tungsten.