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 alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

In a series LCR circuit,the inductive reactance is twice the resistance and the capacitance reactance is ${\frac{1}{3}^{rd}}$ the inductive reactance. The power factor of the circuit is:

  1. $0.5$
  2. $0.6$
  3. $0.8$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l}\omega L = 2R\\frac{1}{{\omega C}} = \frac{1}{3}\left( {\omega L} \right)\\omega L - \frac{1}{{\omega C}} = 2R - \frac{{2R}}{3} = \frac{{4R}}{3}\\tan \phi  = \frac{{4R}}{3} \times \frac{1}{R} = \frac{4}{3}\\cos \phi  = \frac{1}{{\sqrt {1 + {{\tan }^2}\phi } }} = \frac{1}{{\sqrt {1 + \frac{{{4^2}}}{{{3^2}}}} }} = \frac{3}{5} = 0.6\end{array}$

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

An alternative current, L.R circuit comprises of an inductor, whose reactance $X _L = 3R$, where $R$ is the resistance of the circuit. If a capacitor, whose reactance $X _C = R$ is connected in series then what will be the ratio of the new and the old power factor?

  1. $\sqrt{2}$
  2. $\dfrac{1}{\sqrt{2}}$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The power factor of an L-R circuit is cos(phi_old) = R / Z_old, where Z_old = sqrt(R^2 + X_L^2) = sqrt(R^2 + (3R)^2) = R sqrt(10). When a capacitor with X_C = R is added in series, the net reactance becomes X_L - X_C = 3R - R = 2R. The new impedance is Z_new = sqrt(R^2 + (2R)^2) = R sqrt(5). The new power factor is cos(phi_new) = R / Z_new = 1 / sqrt(5). The ratio of new to old power factor is (R / sqrt(5)) / (R / sqrt(10)) = sqrt(10 / 5) = sqrt(2).

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

In an $LR$-circuit, the inductive reactance is equal to the resistance $R$ of the circuit. an e.m.f. $E=E _{0}\ cos(\omega t)$ applied to the circuit. The power consumed in the circuit is

  1. $\dfrac{E^{2} _{0}}{R}$
  2. $\dfrac{E^{2} _{0}}{2R}$
  3. $\dfrac{E^{2} _{0}}{4R}$
  4. $\dfrac{E^{2} _{0}}{8R}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In an LR circuit, Z = sqrt(R^2 + XL^2). Given XL = R, Z = sqrt(2)R. Current amplitude I0 = E0 / Z = E0 / (sqrt(2)R). Average power P = (1/2) * I0^2 * R = (1/2) * (E0^2 / 2R^2) * R = E0^2 / 4R.

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

In an a.c. circuit consisting of resistance $R$ and inductance $L$, the voltage across $R$ is $60$ volt and that across $L $ is $80$ Volt.The total Voltage across the combination is 

  1. $140 V$
  2. $20 V$
  3. $100 V$
  4. $70 V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In an RL series circuit, the total voltage V is the phasor sum of the voltage across the resistor (VR) and the inductor (VL). V = sqrt(VR^2 + VL^2) = sqrt(60^2 + 80^2) = sqrt(3600 + 6400) = sqrt(10000) = 100 V.

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

In an A.C. circuit, the current flowing in inductance is $\displaystyle I=5\sin { \left( 100t-{ \pi  }/{ 2 } \right)  } $ ampers and the potential difference is V = 200 sin (100 t) volts. The power consumption is equal to 

  1. 1000 watt

  2. 40 watt

  3. 20 watt

  4. Zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Power, $\displaystyle P={ I } _{ r.m.s }\times { V } _{ r.m.s }\times \cos { \phi  } $
In the given problem, the phase difference between voltage and current is p/2. Hence
$\displaystyle P={ I } _{ r.m.s }\times { V } _{ r.m.s }\times \cos { \left( { \pi  }/{ 2 } \right)  } =0\ $

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

An inductor $20$ mH, a capacitor $100$ $\mu$F and a resistor $50$ $\Omega$ are connected in series across a source of emf, V$=10$ $\sin 314$t. The power loss in the circuit is?

  1. $2.74$ W
  2. $0.79$ W
  3. $1.13$ W
  4. $0.43$ W
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$L = 20 mH$ $C = 100 \mu F$ $R = 50 \Omega$

$V = 10 sin(314 t)$
$V _0= 10$, $\omega = 314$
$X _L = wL= 314 \times 20\times 10^{-3}= 6.28 \Omega$
$X _C = \dfrac{1}{\omega C}=31.8 \Omega$
$Z = \sqrt{R^2 + (X _C- X _L)^2} = 56.1 $
$Power \ loss P=\dfrac{V _0^2 R}{2 Z^2}= 0.79 W$


Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

Assertion: A resistance is connected to an ac source. Now a capacitor is included in the series circuit. The average power absorbed by the resistance will remain same.  

Reason: By including a capacitor or an inductor in the circuit average power across resistor does not change.

  1. A and R both are true and R is correct explanation of A

  2. A and R both are true but R is not the correct explanation of A

  3. A is true R is false

  4. A is false and R is true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

After connecting the capacitor Irms will charge because impedance is changed.
$\therefore$ A is false

Multiple choice physics communication system commonly used terms in electronic communication system elements of communication system electromagnetic waves and communication system

Thermal noise is generated in:

  1. transistors and diodes

  2. resistors

  3. copper wire

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

Thermal noise or Nyquist noise is the electronic noise generated by the thermal agitation of the charge carriers (usually the electrons) inside an electrical conductor at equilibrium, which happens regardless of any applied voltage.

In each of the options, an electrical element is used and hence this noise is produced in all these elements. The correct option is option D

Multiple choice physics communication system elements of a communication system elements of communication system electromagnetic waves and communication system

Consider telecommunication through optical fibres. Which of the following statement is not true?

  1. Optical fibres may have homogenous core with a suitable cladding.

  2. Optical fibres can be graded refractive index.

  3. Optical fibres are subject to electromagnetic interference from outside.

  4. Optical fibres have extremely low transmission loss.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Optical fibres are not subjected to electro-magnetic interference from outside.

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

The path of cathode rays in an electric field can be approximated to a circle of radius r. In order to double the radius of the circular path, we must 

  1. reduce the electric field to half

  2. double the electric field

  3. increase electric field four times

  4. reduce electric field to one fourth
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, path of cathode rays is nearly circular. So, electric force must be acting as centripetal force for the cathode rays and it is given by
$ F _e = F _c $
qE = $ m \omega ^2 r$
we can conclude that $E \alpha r$ ,provided all the other quantities are constant.
Thus, to double the radius we should double electric field strength.

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

In a meter bridge experiment, the ratio of the left gap resistance to right gap resistance is $2 : 3$, the balance point from left is?

  1. $60$cm
  2. $50$cm
  3. $40$cm
  4. $20$cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $X$ is the left gap resistance and $R$ is the right gap resistance

$l _1$ be the balance point from left

From meter Bridge principle:-
$\implies \dfrac XR= \dfrac{l _1}{100-l _1}=\dfrac23$

$\implies 200-2l _1= 3l _1 \implies 200=5l _1$

$l _1= 40\ cm $

Hence option $(C)$ is correct

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

In the metre bridge experiment of resistances, the known and unknown resistances are inter-changed. The error so removed is:

  1. end correction

  2. index error

  3. due to temperature effect

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

$ \alpha, \beta$ are the end correction on left and right side.


case 1:- Without interchanging.

$ \dfrac{P}{Q} = \dfrac{R}{S} = \dfrac{X+\alpha+l _{1}P}{(Y+\beta+(100-l _{1})P)}$       ..........( 1 )

case 2:- After interchanging.

$\dfrac{P}{Q} = \dfrac{R}{S} = \dfrac{Y+\propto + l _{2}P}{X + \beta+(100-l _{2})P}$    ............( 2 )

on simplification of eq.  (1) and (2) we get
$X = Y +(l _{2}-l _{1})P$
$\therefore$ By interchanging the end correction is removed.

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

Why is the Wheatstone bridge better than the other methods of measuring resistances?

  1. It does not involve Ohm's law

  2. It is based on Kirchoff's law

  3. It has four resistor arms

  4. It is a null method
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Wheatstone bridge is used to measure the unknown resistance by using null method. i.e, when the bridge is balanced, no current through the galvanometer. Using this null method, we can easily measure the unknown resistance if the other three arm's resistor are given.  

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

In a meter bridge an unknown resistance P is connected in the left gap and a $50 \Omega$ resistance in the right gap. Null point is obtained at x cm from the left end. The unknown resistance now shunted with an equal resistance. Find the value of the resistance in the right gap so that the null point is not shifted.

  1. $60 \Omega$
  2. $38 \Omega$
  3. $25 \Omega$
  4. $50 \Omega$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let P be the initial resistance. The null point condition is P/50 = x/(100-x). When P is shunted with an equal resistance P, the new resistance is P/2. To keep the null point at x, the right resistance R' must satisfy (P/2)/R' = x/(100-x). Comparing the two equations, P/50 = (P/2)/R', which gives R' = 25 ohms.