Tag: electric circuits

Questions Related to electric circuits

What is the work function of tungsten at $1500\ K$ temperature, when a diode valve with a tungsten filament works at $1500\ K$? Assume the work function of tungsten at $0\ K$ is $4.52\ eV$:

physics electric circuits introduction to diodes circuit components junction diode and its voltage current characteristics

  1. $4.71\ eV$

  2. $0.39\ eV$

  3. $8.86\ eV$

  4. $1.25\ eV$

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Correct Option: A

Diodes are used to ___________.

physics electric circuits introduction to diodes circuit components junction diode and its voltage current characteristics

  1. convert DC into AC

  2. convert AC into DC

  3. resistance control

  4. current regulation

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Correct Option: B
Explanation:

Diodes are used as a rectifier. A rectifier is a device made from heavily doped p-n junction diode that is used to convert a.c to d.c.

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

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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

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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 $

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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}$

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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. $1.22\ V$

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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

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