Physics

Current Electricity and Circuits

450 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 physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The current voltage relation of diode is given by $I=\left( { e }^{ 1000V/T }-1 \right)mA$, where the applied voltage $V$ is in kelvin. If a student makes an error measuring $\pm 0.01V$ while measuring the current of $5mA$ at $300k$, what will be the error in the value of current in mA?

  1. $0.2mA$
  2. $0.02mA$
  3. $0.5mA$
  4. $0.05mA$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length
Unaware about the fact that analog ammeters and voltmeters can also have zero error, a student recorded following readings while determine resistance by Ohm's law 
If the ammeter has no zero error, the zero error is the voltmeter is 
 Obs, No. Voltage/V  Current/mA   Obs No.  Voltage/V  Current/mA
 1  1.0  40  4  7.0  160
 2  3.0  80  5  9.0  200
 3  5.0  120      
  1. $-1 V$
  2. $-1.5V$
  3. $0.5V$
  4. $1V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Ohm's law V = IR, the slope of the V-I graph gives the resistance. Plotting the points (1.0, 40), (3.0, 80), (5.0, 120), (7.0, 160), (9.0, 200) shows a linear relationship V = 0.05I - 1.0. When I = 0, V = -1.0V, indicating a zero error of -1V.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

A resistor of $10 k\Omega$ having tolerance 10% is connected in series with another resistor of $20k\Omega$ having tolerance 20%. The tolerance of the combination will be:

  1. 10%

  2. 13%

  3. 30%

  4. 20%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In series effective resistance $=R _S=(10k\Omega \pm 10$%)+$(20k\Omega \pm 20$%)$=(30k\Omega \pm 30$%)
$\therefore$ Tolerance of the combination $=( \pm 30$%)

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

Find the percentage error in specific resistance given by $\displaystyle \rho=\frac{\pi r^{2}R}{l}$ where r is the radius having value $\displaystyle \left ( 0.2\pm 0.02 \right )$ cm, R is the resistance of $\displaystyle \left (60\pm 2 \right )\Omega $ and l is length of $\displaystyle \left ( 150\pm 0.1 \right )$ cm.

  1. 5.85%

  2. 11.7%

  3. 23.4%

  4. 35.1%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Applying logarithm on both sides of the given expression and differentiating, 

we get $\dfrac { \Delta \rho  }{ \rho  } =\pm (2\dfrac { \Delta r }{ r } +\dfrac { \Delta l }{ l } +\dfrac { \Delta R }{ R } )$
Given : $\Delta r$=0.02cm, $\Delta R$=2 ohm, $\Delta l$=0.1cm


Substituting the values in above expression,
$\dfrac { \Delta \rho  }{ \rho  } =\pm (2\dfrac { 0.02 }{ 0.2 } +\dfrac { 0.1 }{ 150 } +\dfrac { 2 }{ 60 } )=0.234=23.4$%

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

Two resistors of resistances $R _1$ = (100 $\pm$ 3) $\Omega$ and $R _2$ = (200 $\pm$ 4) $\Omega$ are connected in parallel. The equivalent resistance of the parallel combination is:

  1. (66.7 $\pm$ 1.8) $\Omega$
  2. (66.7 $\pm$ 4.0) $\Omega$
  3. (66.7 $\pm$ 3.0) $\Omega$
  4. (66.7 $\pm$ 7.0) $\Omega$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $R _1 (100 \pm 3) \Omega; R _2 = (200 \pm 4) \Omega$ The equivalent resistance in parallel combination is

$\displaystyle \frac{1}{R _1} = \frac{1}{R _1} + \frac{1}{R _2}, \frac{1}{R _p} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}, R _p = \frac{200}{3} = 66.7 \Omega$

The error in equivalent resistance is given by
$\displaystyle \frac {\Delta R _p}{R _p^2} = \frac {\Delta R _1}{R _1^2} + \frac {\Delta R _2}{R _2^2}; \Delta R _p = \Delta R _1 (\frac{R _p}{R _1})^2 + \Delta R _2 (\frac {R _p}{R _2})^2 = 3 (\frac {66.7}{100})^2 + 4 (\frac {66.7}{200})^2 = 1.8 \Omega $ 
Hence, the equivalent resistance along with error in parallel combination is (66.7 $\pm$ 1.8)$\Omega$.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

A circular copper disc 10 cm in diameter rotates at 1800 revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction B of 1 Wb $m^2$ is perpendicular to disc. What potential difference is developed between the axis of the disc and the rim ?

  1. 0.023 V

  2. 0.23 V

  3. 23 V

  4. 230 V
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, 

$l = r = 5\, cm = 5 \times 10^{-2} m,$
B = 1 Wb $m^{-2}$

$ \omega = 2 \pi \left( \dfrac{1800}{60} \right) \, rad \, s^{-1} = 60 \pi \, rad \, s^{-1},$

$\epsilon \, = \, \dfrac{1}{2} Bl^2 \omega \, =\, \dfrac{1}{2} \times 1 \times (5 \times 10^{-2})^2 \times 60 \pi = 0.23 V$

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

 An inductor of inductance $100\ mH$ is connected in series with a resistance, a variable capacitance and an AC source of frequency $2.0\ kHz$; The value of the capacitance so that maximum current may be drawn into the circuit. 

  1. 50 nF

  2. 60 nF

  3. 63 nF

  4. 79 nF
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l}{X _L} = Lw = {10^{ - 1}} \times 2\pi  \times 2 \times {10^3}\{X _L} = 4\pi  \times {10^2}\Z = \sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} \i = \dfrac{V}{Z} = \dfrac{V}{{\sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} }}\for,{i _{\max }}\{X _L} = {X _C}\\therefore {X _C} = Lw = \dfrac{1}{{Cw}}\C = \dfrac{1}{{{w^2}L}} = \dfrac{1}{{{{10}^{ - 1}} \times 4{\pi ^2} \times 4 \times {{10}^6}}}\ = \dfrac{{{{10}^{ - 5}}}}{{16{\pi ^2}}} = 63nF\end{array}$

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

$5 \mathrm { mV }$ is induced in a coil, when current in another nearby coil changes by $5 \mathrm { A }$ in $0.1$sec. The mutual inductance between the two coils will be

  1. $0.1 \mathrm { H }$
  2. $0.2 \mathrm { H }$
  3. $0.1 \mathrm { mH }$
  4. $0.2 \mathrm { mH }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The induced emf is given by e = M * (di / dt). Rearranging for mutual inductance M gives M = e / (di / dt). Substituting e = 5 mV = 5 x 10^-3 V and di/dt = 5 A / 0.1 s = 50 A/s yields M = (5 x 10^-3) / 50 = 0.1 x 10^-3 H = 0.1 mH.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

For a current carrying inductor, emf associated in $20mV$. Now, current through it changes from $6A$ to $2A$ in $2s$. The coefficient of mutual inductance is 

  1. $20mH$
  2. $10mH$
  3. $1mH$
  4. $2mH$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle \left | e \right |=L\frac{dI}{dt}$
Here, $\displaystyle e=20mV=20\times 10^{-3}V$
Coefficient of mutual inductance,
$ 20\times 10^{-3}=L\times 2$
$\displaystyle \therefore L=10\times 10^{-3}=10mH$

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

When the current in a coil changes from 8 ampere to 2 ampere in $3 \times 10^{-2}$ second, the e.m.f. induced in the coil is 2 volt. The self inductance of the coil (in millinery) is

  1. 1

  2. 5

  3. 20

  4. 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$E.M.F. = L \dfrac{di}{dt}$


$2 = L \times \dfrac{8-2}{3 \times 10^{-2}}$

L = 1 millinery

Here (A) is correct answer

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

Two different coils have self inductance $L _{1}=8\ mH, L _{2}=2\ mH$. The current in the second coil is also increased at the same constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are $i _{1}, V _{1}$ and $W _{1}$ respectively. Corresponding values for the second coil at the same instant are $i _{2}, V _{2}$ and $W _{2}$ respectively. Then 

  1. $\dfrac{i _{1}}{i _{2}}=\dfrac{1}{4}$
  2. $\dfrac{i _{1}}{i _{2}}=4$
  3. $\dfrac{W _{2}}{W _{1}}=4$
  4. $\dfrac{V _{2}}{V _{1}}=\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

We know

$e=L\dfrac{di}{dt}$
$e\propto L$

So,

$\dfrac{e _1}{e _2}=\dfrac{L _1}{L _2}=\dfrac{8}{2}=\dfrac{4}{1}$

Since $P=el=Constant$

Therefore,

$\dfrac{di _1}{dt}=\dfrac{di _2}{dt}$

$P _1=P _2=P$

$e _1i _1=e _2i _2$

$\therefore \dfrac{i _1}{i _2}=\dfrac{e _2}{e _1}=\dfrac{1}{4}$

Thus, ratio of current is 1:4

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

A $50\ Hz$ ac current of peak value $2$ A flows through one of the pair of coils. If the mutual inductance between the pair of coils is $150\ mH$. then the peak value of voltage induced in the second coil is

  1. $30\pi\ V$
  2. $60\pi\ V$
  3. $15\pi\ V$
  4. $300\pi\ V$
  5. $3\pi\ V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the current flows through coil 1 is, ${{I} _{1}}={{I} _{0}}\sin \omega t$ where, ${{I} _{0}}$is peak value of current.

Magnetic flux linked with coil 2 is ${{\phi } _{2}}=M   {{I} _{1}}=M{{I} _{0}}\sin \omega t$

Emf in coil 2 is

${{\varepsilon } _{2}}=\dfrac{d{{\phi } _{2}}}{dt}=\dfrac{d(M{{I} _{0}}\sin \omega t)}{dt}=M \omega {{I} _{0}}\cos \omega t$

So, peak value of voltage induced in coil 2 is $=M{{I} _{o}}\omega ....(1)$

Given that,

$ \nu =50\,Hz $

$ {{I} _{0}}=2\,A $

$ L=150\,mH $

$ \omega =2\pi \nu =2\pi \times 50 $

Put all values in equation (1)

$ current=150\times {{10}^{-3}}\times 2\times 100\pi  $

$ I=30\pi \,V $

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

When 100 volts d.c. is applied across solenoid a current of 1.0 amp flows in it. When 100 volts a.c. is applied across the same coil, the current drops to 0.5 amp. If the frequency of the a.c. source is 50 Hz the impedance and inductance of the solenoid are

  1. 200 ohm and 0.55 Henry

  2. 100 ohm and 0.86 Henry

  3. 200 ohm and 1.0 Henry

  4. 100 ohm and 0.93 Henry
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From DC, R = V/I = 100/1 = 100 ohm. From AC, Z = V/I = 100/0.5 = 200 ohm. Using Z^2 = R^2 + (2*pi*f*L)^2, we find 200^2 = 100^2 + (2*pi*50*L)^2, which solves to L approximately 0.55 H.