Physics

Current Electricity and Circuits

377 Questions

Current electricity and circuits questions cover resistors, EMF, internal resistance, and power calculations in series and parallel configurations. Solving these builds a strong understanding of electrical principles and circuit analysis. These physics problems are highly relevant for technical and science aptitude tests.

Resistor combinationsPower dissipationEMF and internal resistanceAC circuit analysisOperational amplifiers

Current Electricity and Circuits Questions

Multiple choice conductivity and its types electrochemistry

The specific conductance of a $0.01\ M$ solution of $KCl$ is $0.0014\ ohm^{-1} cm^{-1}$ at $25^{\circ}C$. Its equivalent conductance is____________.

  1. $14$
  2. $140$
  3. $1.4$
  4. $0.14$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\kappa =0.0014\ S{ cm }^{ -1 }\ \Lambda _{ eq }=\cfrac { 1000\times \kappa  }{ C } \ C=0.01M\ { \Lambda  } _{ eq }=\cfrac{1000 \times 0.0014}{0.01}=140$

Multiple choice conductivity and its types electrochemistry

The resistance of a N/10 KCI solution is 245$\Omega $. Calculate the equivalent conductance of the solution if the electrodes in the cell are 4cm apart and each having an area of 7.0sq,cm.

  1. $23.32S{ cm }^{ 2 }{ eq }^{ -1 }$
  2. $23.23S{ cm }^{ 2 }{ eq }^{ -1 }$
  3. $2.332S{ cm }^{ 2 }{ eq }^{ -1 }$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given resistance = 245$\Omega $

Formula for the specific conductance (K) = $ \dfrac{1}{R} $ $\times \dfrac{l}{a}$
= $ \dfrac{1}{245} $ $\times \dfrac{4}{7}$
 K = 2.33 $\times 10^-$$^3$ 
Formula for the euivalent conductance=  K $\times$ $\dfrac{1000}{C}$
= 2.33 $\times 10^-$$^3$ $\times 10000$
= 23.32S $cm^2$ eq $^-$$^1$

Multiple choice conductivity and its types electrochemistry

Equivalent constant of standard $BaSO _{4}$ is $400ohm^{-1}\ cm^{2}$ equiv$^{-1}$ and specific conduction is $8\times 10^{-5}\ ohm^{-1}\ cn^{-1}$. Hence $K _{SP}$ of $BaSO _{4}$ is

  1. $4\times 10^{-8}M^{2}$
  2. $1\times 10^{-8}M^{2}$
  3. $2\times 10^{-4}M^{2}$
  4. $1\times 10^{-4}M^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice conductivity and its types electrochemistry

The resistance of 0.1 N solution of a salt is found to be $2.5\times10^{3}$. The equivalent conductance of the solution is: (cell constant=1.15 $cm^{-1}$)

  1. 3.6

  2. 4.6

  3. 5.6

  4. 6.6
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relationship between the specific conductance, resistance and the cell constant is $ \kappa =\cfrac { 1 }{ R } \times \cfrac { l }{ a }$.
Substitute $ R=2.5\times { 10 }^{ 3 }\quad ohm $ and $ \cfrac { l }{ a } =1.15\quad {cm }^{ -1 } $.
Hence $ \kappa =\cfrac { 1 }{ 2.5\times { 10 }^{ 3 }\quad ohm } \times1.15\quad { cm }^{ -1 }=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
The relationship between the equivalent conductance and specifc conductance is $  { \Lambda  } _{ eq }=\cfrac { \kappa \times 1000 }{ M }  $.
Substitute $ M=0.1\quad N $ and $ \kappa=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
Hence $ { \Lambda  } _{ eq }=\cfrac { \cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \times 1000 }{ 0.1 } \quad =\quad 4.6\quad \quad { ohm }^{ -1 }\quad { cm }^{ 2 }\quad { equiv }^{ -1 } $.

Multiple choice conductivity and its types electrochemistry

The resistance of $N/10$ solution is found to be $2.5\times 10^{3}ohm$. The equivalent conductance of the solution is (cell constant $= 1.25\ cm^{-1})$.

  1. $2.5\ ohm^{-1} cm^{2} equiv^{-1}$
  2. $5\ ohm^{-1} cm^{2} equiv^{-1}$
  3. $2.5\ ohm^{-1} cm^{-2} equiv^{-1}$
  4. $5\ ohm^{-1} cm^{-2} equiv^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given Data : 1)  Resistance = $R$ = $2.5 \times 10^3$

                     2) Cell constant = $k$ = $\cfrac{l}{a}$= $1.15$$cm^2$
                     3)  Normality =$N$= $0.1 N$
To Find : Equivalent conductance = $\Lambda$$ _e$$ _q$

The relation between $K$ ,$R$ and $ k$ is,
$K$ = $\cfrac{1}{R} \times \cfrac{l}{a}$
where $K$ is specific conductance.
$\therefore$ Substituting the given values we get,

$K$ = $\cfrac{1}{2.5×10^3}$ × $1.15cm^-$$^1$

     = $\cfrac{1.15}{2.5×10^3}$ $ohm$$^-$$^1$$cm$$^-$$^1$

The relation between $\Lambda$$ _e$$ _q$ and $K$ is,

$\Lambda$$ _e$$ _q$ = $\cfrac{K×1000}{N}$

Substituting the value of $M$ and $K$ we get,

$\Lambda _{eq}= \cfrac{1.15}{2.5\times10^3\times0.1} \times 1000$

       = $4.6$$ohm^-$$^1$$cm$$^2$$equi$$^{-1}$      [Note:$ \text {Normality= no. of equiv.} /cm^3$]

Here appproximation is taken,

      $\approx$ $5$ $ohm$$^-$$^1$$cm$$^2$$equi$$^-$$^1$

Hence the correct option is 'B'.

Multiple choice conductivity and its types electrochemistry

The resistance of $0.01\ N$ solution at $25^{\circ}$ is $200\ ohm$. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

  1. $200\ ohm^{-1}cm^{2} eq^{-1}$.
  2. $300\ ohm^{-1}cm^{2} eq^{-1}$.
  3. $400\ ohm^{-1}cm^{2} eq^{-1}$.
  4. $500\ ohm^{-1}cm^{2} eq^{-1}$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

since, we have

$conductance * cell constant$ =  specific conductance
cell consyant = 1
conductance = specific conductance= $\dfrac{1}{200}$

equivalent conductance= $\dfrac{K*1000}{N}$ $Scm^{2}eq^{-1}$

equivalent conductance= $\dfrac{1*1000}{200*0.01}$

equivalent conductance= $500$ $ Scm^{2}eq^{-1}$
 

Multiple choice conductivity and its types electrochemistry

If the specific resistance of a solution of concentration C g equivalent/litre is R, then its equivalent conductance is:

  1. $\dfrac{100R}{C}$
  2. $\dfrac{RC}{1000}$
  3. $\dfrac{1000}{RC}$
  4. $\dfrac{C}{1000R}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Specific resistance for $C _g eq/lt=R$

Dont know the meaning of conductance it must be conductance
Conductance of solution $=k=\dfrac{1}{k}$
Equivalent conductance $=\dfrac{k\propto 1000}{c}$
                                         $=\dfrac{1000}{RC}$

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

Alternating current is flowing in inductance L and resistance R. The frequency of source is $\displaystyle\frac{\omega}{2\pi}$. Which of the following statement is correct.

  1. For low frequency the limiting value of impedance is L
  2. For high frequency the limiting value of impedance is $L\omega$

  3. For high frequency the limiting value of impedance is R
  4. For low frequency the limiting value of impedance is $L\omega$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} \, \, As\, frequency\, approaches\, zero\, or\, DC,\, the\, inducators\, reac\tan  ce\, would\, decrease\, tozero\, , \ acting\, like\, a\, short\, circuit.\, this\, means\, inductive\, reac\tan  ce\, is\, proportional\, to\, fequency \ \, \, \, \, \, \, \, \, \, \, \, \, \, so\, ,\, for\, low\, frequency\, the\, { { limimiting } }\, \, value\, of\, impedance\, is\, L,\, and\, \, alternating\,  \ current\, is\, flowing\, in\, inductance\, L\, and\, resistance\, R.\, \, The\, frequency\, of\, source\, is\, \frac { \omega  }{ { 2\pi  } } . \ so\, the\, correct\, option\, is\, A. \end{array}$

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

A student measures the terminal potential difference (V) of a cell (of emf $\varepsilon$ and internal) resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively equal to :



  1. $-\in \;and\;r$
  2. $\in \;and\;-r$
  3. $-r\;and\;\in \;$
  4. $r\;and\;-\in$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$E = V + Ir $
$\Rightarrow V=E-Ir$ 
$Comparing\;with\; y = mx + c$ 
$Slope = - r, intercept = E$




Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

A source of 220 V is applied in an A C circuit . The value of resistance is 220 $\Omega$. Frequency & inductance are 50Hz & 0.7 H then wattless current is 

  1. 0.5 amp

  2. 0.7 amp

  3. 1.0 amp

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A source= $220V$

The value of resistance= $220 \Omega$
Frequency= $50 Hz$
Inductance= $0.7H$
Find the wattless current= ?
Wattless component of current is $i=i _v\sin \theta$
                                                            $=\cfrac {Ev}{z}\sin \theta$
where, $z=$ impedance of $L-R$ circuit
                $=\sqrt {R^2+L^2W^2}$ so,
$i=\cfrac {220}{\sqrt {R^2+L^2+W^2}}\sin \theta$ from impedance triangle,
$\sin \theta= \cfrac {LW}{\sqrt {R^2+L^2W^2}}$
$\Rightarrow i=\cfrac {220}{\sqrt {R^2+L^2W^2}}\cfrac {LW}{\sqrt {R^2+L^2W^2}}$
        $=\cfrac {220}{R^2+L^2W^2}LW$
        $=\cfrac {220 \times 0.7 \times 2 \Pi \times 50}{(220)^2+(0.7\times 2\Pi \times 50)^2}$
        $=\cfrac {220 \times 220}{(220)^2+(220)^2}$
        $=0.5 A$ .

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

The time constant of a circuit is 10 sec, When a resistance of $ 100 \Omega $ is connected in series in a previous circuit then time constant becomes 2 second,then the self inductance of the circuit is;-

  1. $250 H$
  2. $50H$
  3. $150 H$
  4. $25 H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In LR circuit,

The time constant $\tau=\dfrac{L}{R}$
$10=\dfrac{L}{R}$
$L=10R$. . . . . . .(1)
When Resistance $100\Omega $ is connect in series, than the time constant is
$\tau'=\dfrac{L}{R+100}=2s$
$L=2R+200$. . . . . . .(2)
Equating equation (1 ) and (2), we get
$2R+200=10R$
$8R=200$
$R=25\Omega$
From equation (1),
$L=10R=10\times 25$
$L=250H$

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

if the length and area of cross section of an inductor remain same but the number of turns is doubled its self inductance will become:

  1. half

  2. four time

  3. double

  4. one- fourth
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Self-inductance L is proportional to the square of the total number of turns (N^2). If N is doubled, L becomes 2^2 = 4 times the original.