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 electric circuits combination of resistors combination of cells combinations of components

A battery is delivering same power to resistance $R _1$ and $R _2$ . Then find the internal resistance of battery :

  1. $\dfrac { R _ { 1 } - R _ { 2 } } { 2 }$
  2. ${ R _ { 1 } + R _ { 2 } }$
  3. $ \sqrt { R _ { 1 } + R _ { 2 } }$
  4. $\sqrt { R _ { 1 } + R _ { 2 } / 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\begin{array}{l} P=\frac { { { V^{ 2 } } } }{ { { R^{ 2 } }+r } } -------\left( 1 \right)  \\ P=\frac { { { V^{ 2 } } } }{ { { R^{ 2 } }-r } } -------\left( 2 \right)  \\ solve\, \, eqation\left( 1 \right) and\left( 2 \right)  \\ =\dfrac { { { R _{ 1 } }-{ R _{ 2 } } } }{ 2 }  \end{array}$
Hence,
option $(A)$ is correct answer.
Multiple choice physics electric current and its effects electric bell the magnetic effect of a current current electricity and magnetism

An electric bell has a resistance of $5 \Omega$ and requires a current of $0.25\ A$ to work it. Assuming that the resistance of the bell wire is $1 \Omega$ per $15m$ and that the bell push is $90m$ distance from the bell. How many cells each of emf $1.4V$ and internal resistance $2 \Omega$, will be required to work the circuit.

  1. $3$
  2. $4$
  3. $5$
  4. $Can't\ be\ determined$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total resistance = R_bell + R_wire = 5 + (90/15)*1 = 11 ohms. Current = 0.25A. Total EMF required = I * R_total = 0.25 * (11 + n*r_internal). Solving for n with 1.4V cells.

Multiple choice safety precautions to be taken while handling electricity ohm's law electric charge and electric current current electricity physics

Ohm's law in vector form is :

  1. $V=I.R$
  2. $\overrightarrow { J } =\sigma \overrightarrow { E } $
  3. $\overrightarrow { J } =\rho \overrightarrow { E } $
  4. $\overrightarrow { E } =\sigma \overrightarrow { J } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ohm's law in vector form relates current density J to electric field E via conductivity sigma: J = sigma * E.

Multiple choice physics kirchhoff's law circuit problems kirchoff's law and problems on it equivalent resistance in series and parallel connection

Between any two points in a circuit, the sum of all .............. is the same through any pathway.

  1. charge

  2. current

  3. potential difference

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

Kirchhoff's voltage law says that if we add up all the voltage drops that occur around the loop, the total will be equal to the total voltage applied across the loop.hence between any two point in the circuit , the sum of all the voltage drops is the same through any pathway.

Multiple choice physics kirchhoff's law circuit problems kirchoff's law and problems on it equivalent resistance in series and parallel connection

For a $DC$ circuit, Kirchoff's rules yield the following equations.
$I _{3}=I _{1}+I _{2}$
$10 = 3I _{1}-2I _{2}$
$50=2I _{2}+9.6I _{3}$
What is the current $I _{2}$ (Amps)?

  1. $0.131$
  2. $-1.37$
  3. $0.245$
  4. $-3.5$
  5. $1.00$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Given: 1) I3 = I1 + I2, 2) 10 = 3I1 - 2I2, 3) 50 = 2I2 + 9.6I3. Substitute (1) into (3): 50 = 2I2 + 9.6(I1 + I2) = 9.6I1 + 11.6I2. Now solve the system: 3I1 - 2I2 = 10 (multiply by 3.2: 9.6I1 - 6.4I2 = 32). Subtract from 9.6I1 + 11.6I2 = 50: (11.6 + 6.4)I2 = 50 - 32 => 18I2 = 18 => I2 = 1.0 A.

Multiple choice physics kirchhoff's law combination of resistors combination of cells combinations of components

The diagram shows a $40\Omega$ resistor and a $60\Omega$ resistor connected in parallel.
What is the total resistance between points P and Q?

  1. less than $40\Omega$
  2. $50\Omega$
  3. between $50\Omega$ and $100\Omega$
  4. $100\Omega$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
For parallel connection
$R _{eq}= \dfrac{R _1 \times R _2}{R _1 +R _2}$
$\implies R _{eq}= \dfrac{40 \times 60}{40+60}$
$\implies R _{eq}= 24 \Omega$

So, it is less than $40 \Omega$
Multiple choice capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

The inductance of the oscillatory circuit of a radio station is 10 milli henry band its capacitance is $0.25 \mu F$. Taking the effect of the resistance negligible, wavelength of the broadcasted waves will be (velocity of light = $3.0  \ 10 ^4 \ m/s, \pi = 3.14$):

  1. $9,42 \times 10^4 m$
  2. $18.8 \times 10^4 m$
  3. $4.5 \times 10^4 m$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
C 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.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
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

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 } }  } $