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 ac and dc ac vs dc electro-magnetism effects of electric current physics

The resistance of a coil for dc is $R$ ohms. When connected to ac supply the total resistance offered by circuit to ac current.

  1. Will remain same

  2. Will increase

  3. Will decrease

  4. Will be zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Resistance offered by coil is same for $AC$ as well as $DC.$ So, Resistance will not change.

Multiple choice ac and dc ac vs dc electro-magnetism effects of electric current physics

Assertion: Ohm's law cannot be applied to a.c circuit. 
Reason: Resistance offered by capacitor for a.c source depends upon the frequency of the source. 

  1. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

  2. If both Assertion and Reason are true but the Reason is not the correct explanation of the Assertion.

  3. If Assertion is true statement but Reason is false.

  4. If both Assertion and Reason are false statements.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Assertion statement is false.
Ohm's can also be applied to a.c circuit.
Reason statement is correct.
Capacitor resistance, $R _C =\dfrac{1}{wC} = \dfrac{1}{(2\pi \nu)C}$
Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

An electrical device draws 2 kW power from ac mains voltage 223 V(rms). The current differs lags in phase by $\phi = tan^{-1} \left ( -\frac{3}{4} \right )$ as compared to voltage. The resistance R in the circuit is:

  1. 15 $\Omega$
  2. 20 $\Omega$
  3. 25 $\Omega$
  4. 30 $\Omega$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $P \, = \, 2 \, kW \, = \, 2 \, \times \, 10^3W$
$V _{rms} \, = \, 233 \, V, tan \, \phi \, = \, -\dfrac{3}{4}$
$As, \, P \, = \, \dfrac{V^2 _{rms}}{Z}$

$\Rightarrow \, Z \, = \, \dfrac{V^2 _{rms}}{P} \, = \, \dfrac{(223)^2}{2000} \, =  \dfrac{49729}{2000} \, = \, 24.86 \, \Omega \, or \, Z \, = \, 25 \, \Omega$

$\tan \, \phi \, = \, \dfrac{X _C \, - \, X _L}{R} \, = \, - \dfrac{3}{4} \, \therefore \, X _C \, - \, X _L \, = \, -\dfrac{3}{4} R.$

AS, $Z^2 \, = \, R^2 \, + \, (X _C \, - \, X _L)^2$
$\therefore \, (25)^2 \, = \, R^2 \, + \, \left(-\dfrac{3}{4} \, R \right)^2$
$625 \, = \, \dfrac{25 \, R^2}{16}.$
$R^2 \, = \, \dfrac{625 \, \times \, 16}{25} \,  \, \Rightarrow \, R \, = \, 20 \, \Omega$

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

A coil has a resistance $ 10 \Omega $ and an inductance of 0.4 henry. It is connected to an AC source of $ 6.5 V , \frac {30} { \pi } Hz. $ The average power consumed in the circuit, is :

  1. $ \cfrac {5} { 8} W $
  2. $ \cfrac {4} {3} W $
  3. $ \cfrac {3} {8} W $
  4. $ \cfrac {6} {7} W $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The impedance Z = sqrt(R^2 + XL^2). XL = 2 * pi * f * L = 2 * pi * (30/pi) * 0.4 = 24 ohms. Z = sqrt(10^2 + 24^2) = sqrt(100 + 576) = 26 ohms. Current I = V/Z = 6.5 / 26 = 0.25 A. Power P = I^2 * R = (0.25)^2 * 10 = 0.0625 * 10 = 0.625 W, which is 5/8 W.

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
B Correct answer
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 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.