Tag: electric circuits

Quizzes (8) Questions (30)

Questions Related to electric circuits

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

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 $4.52\ eV$:

  1. $4.71\ eV$
  2. $0.39\ eV$
  3. $8.86\ eV$
  4. $1.25\ eV$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The work function at temperature T is given by phi(T) = phi(0) + 2kT. Given phi(0) = 4.52 eV, k = 8.617 * 10^-5 eV/K, and T = 1500 K, the increase is 2 * 8.617 * 10^-5 * 1500 = 0.2585 eV. 4.52 + 0.26 = 4.78 eV. Option A is the closest.

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

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?

  1. $3$A
  2. $6$A
  3. $2$A
  4. $1$A
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $1+2=3A$

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

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:

  1. 1.2

  2. 2

  3. 4

  4. 4.8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For identical cells, the current in series is E/(R + nr) and in parallel is E/(R + r/n). Setting these equal for R=4 gives r=4.

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

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

  1. 4$\Omega $
  2. 6$\Omega $
  3. 8$\Omega $
  4. 10$\Omega $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ \Rightarrow  (100-12) = I[3+R]$

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

$ \Rightarrow  R = 8\Omega$

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

$n$ 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 :

  1. $R = nr$
  2. $R = r$
  3. $r = nR$
  4. $ R = \sqrt {n} \dfrac{E}{r}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For cells in series, current is E / (R + nbr), and in parallel, current is E / (R + r/n). Equating the currents under the condition that external resistance is the same gives R = r. Thus, the external resistance must equal the internal resistance of a single cell.

Multiple choice physics electric circuits combination of resistors combination of cells 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

  1. $1.12\ V$
  2. $1.04\ V$
  3. $1.18\ V$
  4. $1.22\ V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the potentiometer principle, the potential drop across 26 cm of the wire is proportional to the EMF of the cell. Calculation: (2.2 * 26/50) / (1 + 0.1) * (1) = 1.22V.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $\frac {r _1-r _2}{2}$
  2. $\frac {r _1+r _2}{2}$
  3. $r _1-r _2$
  4. $r _1+r _2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $V _1=0 \Rightarrow E-\dfrac{2Er _1}{r _1+r _2+R}=0$ 

$r _1+r _2+R-2r _1=0$ 

$R=r _1-r _2$

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

$24$ 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 :

  1. In series

  2. in parallel
  3. In $4$ rows, each row having $6$ cells
  4. In $6$ rows, each row having $4$ cells
Reveal answer Fill a bubble to check yourself
A Correct answer
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