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 effects of electric current joule's law of heating thermal effect of electric current heating effect of electric current

Laws of heating are given by

  1. Joule

  2. Ohm

  3. Maxwell

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

Laws of heating are given by Joule 

Joule's Heating Effect:- When an electric current is passed through the conductor, it becomes hot and its temperature starts rising. This is known as the heating effect of current or Joule's heating effect.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A.c across L-R,L-C and L-C-R series circuits. In an LR circuit, $R=10\Omega$ and $L=2H$, If an alternating voltage of $120V$ and $60Hz$ is connected in this circuit, then the value of current flowing in it will be _____ A (nearly)

  1. $0.32$
  2. $0.16$
  3. $0.48$
  4. $0.8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Impedance Z = sqrt(R^2 + (2*pi*f*L)^2). R = 10, L = 2, f = 60. XL = 2 * 3.14 * 60 * 2 = 753.6. Z = sqrt(100 + 567913) approx 754. Current I = V/Z = 120 / 754 approx 0.16 A.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

An L-C-R series circuit with $100\omega$ resistance is connected to an A.C source of 200 V and angular frequency $300 rad\,s^{-1}$. When only the capacitor is removed, the current lags behind the voltage by $60^0$ . When only inductor is removed, the current leads the voltage by $60^0$. If all elements are connected , the current in the circuit is

  1. 0.5 A

  2. 1.5 A

  3. 2 A

  4. 2.5 A
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given R = 100. When C is removed (LR circuit), tan(60) = XL/R => XL = 100 * sqrt(3) = 173.2. When L is removed (RC circuit), tan(60) = XC/R => XC = 100 * sqrt(3) = 173.2. Since XL = XC, the circuit is at resonance when all elements are connected. At resonance, Z = R = 100. Current I = V/R = 200 / 100 = 2 A.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A coil has self-inductance $L = 0.04\, H$ and resistance $R = 12 \Omega$ , connected to $220 V$, 50 Hz supply, what will be the current flow in the coil ?

  1. 11.7 A

  2. 12.7 A

  3. 10.7 A

  4. 14.7 A
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $L = 0.04 \, H, R= 12\Omega$
$V= 220$ volt and $f= 50 Hz$
The value of current
$I=\dfrac {V}{Z}$
or or $ I=\dfrac {V}{\sqrt{R^2+(\omega L)^2}}$
or $I=\dfrac {V}{\sqrt{R^2+(2\pi fL)^2}}$
or $I=\dfrac {220}{\sqrt{144+(2\pi50\times 0.04)^2}}$
$\Rightarrow I= 12.7\, A$

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

What is the rms value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of $2$ amperes in the same resistor:

  1. $6$ amp
  2. $2$ amp
  3. $3.46$ amp
  4. $0.66$ amp
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The heat produced in a resistor is proportional to the square of the current (I^2 * R * t). Since the alternating current produces thrice the heat of a direct current of 2 amperes, we have I_rms^2 = 3 * I_dc^2 = 3 * (2^2) = 12. Taking the square root of 12 gives approximately 3.46 amperes.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

When $100$ volt D.C is applied across a coil, a current of one ampere flows through it, when $100V$ ac of $50Hz$ is applied to the same coil, only $0.5amp$ flows. Calculate the resistance and inductance of the coil.

  1. $300\Omega ,\left( \sqrt { 3 } /\pi \right) Hz$
  2. $100\Omega ,\left( \sqrt { 3 } /\pi \right) Hz$
  3. $200\Omega ,\left( \sqrt { 3 } /\pi \right) Hz$
  4. $400\Omega ,\left( \sqrt { 3 } /\pi \right) Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For DC, R = V/I = 100/1 = 100 ohms. For AC, Z = V/I = 100/0.5 = 200 ohms. Since Z^2 = R^2 + (wL)^2, then 200^2 = 100^2 + (2*pi*50*L)^2. Solving for L gives sqrt(30000)/(100*pi) = sqrt(3)/pi H. The unit Hz in the option is a typo for H.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

An alternating current of $1.5mA$ and angular frequency $\omega=300rad/s$ flows through $10k\Omega$ resistor and a $0.50\mu F$ capacitor in series. Find the RMS voltage across the capacitor and impedance of the circuit?

  1. $20V,12\Omega$
  2. $10V,12\Omega$
  3. $10V,13\Omega$
  4. $40V,12\Omega$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Voltage across capacitor V_c = I_rms * X_c = I_rms * (1/(w*C)). Given I_rms = 1.5mA, w = 300, C = 0.5uF, V_c = 0.0015 / (300 * 0.5 * 10^-6) = 0.0015 / 0.00015 = 10V. Impedance Z = sqrt(R^2 + X_c^2). With R = 10k ohms and X_c = 1/(300 * 0.5 * 10^-6) = 6666 ohms, Z is approx 12k ohms. Option B is the closest match.

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A sinusoidal voltage ${ V } _{ 0 }\sin { \omega t } $ is applied across a series combination of resistance R and inductor L. The amplitude of the current in the circuit is :

  1. $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }+{ \omega }^{ 2 }{ L }^{ 2 } } } $
  2. $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }-{ \omega }^{ 2 }{ L }^{ 2 } } } $
  3. $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }+{ \omega }^{ 2 }{ L }^{ 2 } } } \sin { \omega t } \quad $
  4. ${ V } _{ 0 }/R$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Impedance of the circuit $\sqrt { { R }^{ 2 }+{ \omega  }^{ 2 }{ L }^{ 2 } } $
Amplitude of voltage$={V} _{0}$
$\therefore$ Amplitude of current $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }-{ \omega  }^{ 2 }{ L }^{ 2 } }  } $

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

An ideal choke takes a current of $8A$ when connected to an a.c source of $100volt$ and $50Hz$. A pure resistor under the same conditions takes a current of $10A$. If two are connected in series to an a.c supply of $100V$ and $40Hz$, then the current in the series combination of above resistor and inductor is :

  1. $10A$
  2. $8A$
  3. $5\sqrt{2}$ amp
  4. $10\sqrt {2}$ amp
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ X } _{ L }=\cfrac { 100 }{ 8 } ;R=\cfrac { 100 }{ 10 } =10\Omega $
$L\times 100\pi =\cfrac { 100 }{ 8 } $
$L=\cfrac { 1 }{ 8\pi  } H$
$Z=\sqrt { { \left( \cfrac { 1 }{ 8\pi  } \times 2\pi \times 40 \right)  }^{ 2 }+{ 10 }^{ 2 } } =10\sqrt { 2 } $
$I=\cfrac { E }{ Z } =\cfrac { 100 }{ 10\sqrt { 2 }  } =\cfrac { 10 }{ \sqrt { 2 }  } =5\sqrt { 2 } A$

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A coil of negligible resistance is connected in series with $90\Omega$ resistor across a $120V-60Hz$ line. A voltmenter reads $36V$ across the resistance. Find the voltage across the coil and inductance of the coil.

  1. $114V,1.76H$
  2. $114.5V,0.76H$
  3. $114V,0.86H$
  4. $144V,0.76H$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$V=\sqrt{V _R^2+V _L^2}\ \therefore V _L=\sqrt{V^2-V _R^2}\=114.5V\V _R=IR\ \Rightarrow I+\cfrac{36}{90}=0.4A\ \therefore V _L=IX _L\=I\omega L\ \therefore L=\cfrac{V _L}{I\omega}=\cfrac{114.5}{0.4\times2\pi\times60}\=0.76H$

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A $200km$ long telegraph wire has capacity of $0.014\mu F/km$. If it carries an alternating current of $50KHz$, what should be the value of an inductance required to be connected in series so that impedance is minimum?

  1. $0.703H$
  2. $0.303H$
  3. $0.503H$
  4. $0.603H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A $0.19H$ inductor and a $80\Omega$ resistance connected in series to a $220V, 50Hz$ ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

  1. $2.2A, tan^{-1} ({3 \over 4})$
  2. $3A, tan^{-1} ({2 \over 5})$
  3. $5A, tan^{-1} ({8 \over 9})$
  4. $6A, tan^{-1} ({7 \over 5})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Z = sqrt(R^2 + (wL)^2). w = 2*pi*50 = 314 rad/s. wL = 314 * 0.19 approx 60 ohms. Z = sqrt(80^2 + 60^2) = 100 ohms. I = V/Z = 220/100 = 2.2A. Phase angle tan(phi) = wL/R = 60/80 = 3/4. Thus phi = tan^-1(3/4).

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A circuit containing an inductance and a resistance connected in series, has an AC source of $200V$, $50Hz$ connected across it. An AC current of $10A$ rms flows through the circuit and the power loss is measured to be $1kW$. Find
(a) the inductance in the circuit
(b) the frequency of the AC when the phase difference between the current and emf becomes $\pi /4$. with the above components.

  1. (a) $\cfrac { \sqrt { 3 } }{ 70\pi } H$ (b) $\cfrac { 50 }{ \sqrt { 3 } } Hz$
  2. (a) $\cfrac { \sqrt { 3 } }{ 10\pi } H$ (b) $\cfrac { 50 }{ \sqrt { 3 } } Hz$
  3. (a) $\cfrac { \sqrt { 3 } }{ 20\pi } H$ (b) $\cfrac { 50 }{ \sqrt { 4 } } Hz$
  4. (a) $\cfrac { \sqrt { 3 } }{ 60\pi } H$ (b) $\cfrac { 50 }{ \sqrt { 3 } } Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power P = I^2 * R. 1000 = 100 * R, so R = 10 ohms. Z = V/I = 200/10 = 20 ohms. Z^2 = R^2 + (wL)^2, so 400 = 100 + (2*pi*50*L)^2, 300 = (100*pi*L)^2, L = sqrt(3)/(10*pi). For phase angle pi/4, tan(pi/4) = 1 = w'L/R, so w' = R/L = 10 / (sqrt(3)/(10*pi)) = 100*pi/sqrt(3). f' = w'/(2*pi) = 50/sqrt(3).

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A solenoid of 10 Henry inductance and 2 ohm resistance, is connected to a 10 volt battery. In how much time the magnetic energy will be reaches to 1/4th of the maximum value?

  1. 3.5 sec

  2. 2.5 sec

  3. 5.5 sec

  4. 7.5 sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

L = 10 H

R = 2 ohm

V = 10 volt

Now, the maximum current is

  $ {{i} _{0}}=\dfrac{V}{R} $

 $ {{i} _{0}}=\dfrac{10}{2} $

 $ {{i} _{0}}=5A $

The maximum energy is

  $ {{E} _{0}}=\dfrac{1}{2}Li _{0}^{2} $

 $ {{E} _{0}}=\dfrac{1}{2}\times 10\times 5\times 5 $

 $ {{E} _{0}}=125\,J $

Now, the magnetic energy

  $ E=\dfrac{{{E} _{0}}}{4} $

 $ E=\dfrac{125}{4}\,J $

Now, 

  $ E=\dfrac{1}{2}L{{i}^{2}} $

 $ \dfrac{125}{4}=\dfrac{1}{2}L{{i}^{2}} $

 $ \dfrac{125}{2\times 10}={{i}^{2}} $

 $ {{i}^{2}}=\dfrac{25}{2} $

 $ i=\dfrac{5}{2} $

 $ i=2.5\,A $

Now, the time taken to rise current from 0 - 2.5A.

We know that, the instantaneous current during its growth in an L-R circuit.

 $ i={{i} _{0}}\left( 1-{{e}^{-\frac{Rt}{L}}} \right) $

 $ 2.5=5\left( 1-{{e}^{-\frac{Rt}{L}}} \right) $

 $ {{e}^{\frac{-Rt}{L}}}=0.5 $

 $ \dfrac{-Rt}{L}=\ln (0.5) $

 $ \dfrac{Rt}{L}=0.693 $

 $ t=\dfrac{6.93}{2} $

 $ t=3.46 $

 $ t=3.5\sec  $

Hence, the magnetic energy will be increases to $\dfrac{1}{4}$ of the maximum value at $3.5$ sec.

 

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A coil of inductance $8.4\ mil$ and resistance $6W$ is connected to a $12V$ battery. The current in the coil is $1.0\ A$ at approximately the time

  1. $500\ s$
  2. $25\ $s
  3. $35\ s$
  4. $1\ ms$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The time constant tau = L/R = 8.4mH / 6 ohms = 1.4ms. The current in an RL circuit is I(t) = (V/R)(1 - e^(-t/tau)). Here V/R = 12/6 = 2A. We want I(t) = 1A, so 1 = 2(1 - e^(-t/tau)), 0.5 = 1 - e^(-t/tau), e^(-t/tau) = 0.5. t = tau * ln(2) = 1.4ms * 0.693 approx 1ms.