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 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

 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

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.

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

Two resistances $r _1=(5.0\pm 0.2)\Omega$ and $r _2=(10.0\pm 0.1)\Omega$ are connected in parallel. Find the value of equivalent resistance with limits of percentage error.

  1. $r _p=4\Omega\pm $7%
  2. $r _p=3.3\Omega\pm $14%
  3. $r _p=3.3\Omega\pm $3.5%
  4. $r _p=3.3\Omega\pm $7%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $r _1=5 \Omega , r _2=10 \Omega, \Delta r _1=0.2 \Omega $ and $\Delta r _2=0.1 \Omega$


The equivalent resistance, $r _p=\dfrac{r _1r _2}{r _1+r _2}=\dfrac{5\times 10}{5+10}=3.3 \Omega$

$\dfrac{\Delta r _p}{r _p} = \dfrac{\Delta r _1}{r _1}+\dfrac{\Delta r _2}{r _2}+\dfrac{\Delta (r _1+r _2)}{r _1+r _2}=\dfrac{0.2}{5}+\dfrac{0.1}{10}+\dfrac{0.2+0.1}{5+10}=0.07$

Thus, The value of equivalent resistance with limits of  % error $=r _p\pm\Delta r _p=3.3 \Omega \pm 7 \%$

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

A current of $5A$ passes through a copper conductor $(resistivity=1.7\times10^{-8}\Omega m$) of radius of cross-section $5mm$. Find the mobility of the charges if their drift velocity is $1.1\times 10^{-3}m/s$

  1. $1.3m^2/Vs$
  2. $1.5m^2/Vs$
  3. $1.8m^2/Vs$
  4. $1.0m^2/Vs$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\mu=\dfrac{V _d}{E}$     $E=\rho J$


$=\dfrac{1.1\times 10^{-3}}{1.7\times 10^{-8}\times \dfrac{5}{\pi \times 25\times 10^{-6}}}$

$=\dfrac{1.1\times 10^{-3}\times \pi \times 25\times10^{-6}}{1.7 \times 10^{-8} \times5}\approx 1.01m^2/Vs$

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

The electric current in a wire may be calculated using the equation $I=Anvq$.
Which statement is not correct?

  1. n is the number of charge carriers per unit volume of the wire

  2. nA is the number of charge carriers per unit length of the wire

  3. q is the charge of each charge carrier

  4. v is the velocity of each charge carrier
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,

Electric current in the wire  , $I=Anvq$

Here $A$ is the cross-section of wire 
$n$ is the number of charge carriers per unit volume of wire
$v $ is  drift velocity of charge carriers 
$q$ is the charge of each charge carier 

Hence incorrect statement is $(D)$

Multiple choice

What is the resting membrane potential of a neuron?

  1. -70 mV

  2. -50 mV

  3. -30 mV

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

The resting membrane potential of a neuron is typically around -70 mV, which is the difference in electrical potential between the inside and outside of the neuron.

Multiple choice

In an electrical circuit, the equation (L\frac{di}{dt} + Ri = E) describes the current (i) flowing through an inductor with inductance (L), a resistor with resistance (R), and a voltage source (E). What is the time constant of the circuit?

  1. \(\frac{L}{R}\)
  2. \(\frac{R}{L}\)
  3. \(\frac{E}{R}\)
  4. \(\frac{E}{L}\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The time constant of the circuit is given by (\tau = \frac{L}{R}).

Multiple choice

What is the relationship between voltage, current, and resistance in an electric circuit?

  1. Ohm's Law

  2. Kirchhoff's Current Law

  3. Kirchhoff's Voltage Law

  4. Faraday's Law
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ohm's Law states that the voltage across a conductor is directly proportional to the current flowing through it, provided the temperature and other physical conditions remain constant.

Multiple choice

What is the impedance of a transducer?

  1. The resistance of the transducer to the flow of electrical current

  2. The capacitance of the transducer

  3. The inductance of the transducer

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The impedance of a transducer is a measure of all of the above. It includes the resistance of the transducer to the flow of electrical current, the capacitance of the transducer, and the inductance of the transducer.