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 effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

A resistor has resistance R. When the potential difference across the resistor is V, the current in
the resistor is I. The power dissipated in the resistor is P. Work W is done when charge Q flows
through the resistor.
What is not a valid relationship between these variables? 

  1. $I =\frac {P}{V}$
  2. $Q =\frac {W}{V}$
  3. $R =\frac {P}{I^2}$
  4. $R =\frac {V}{P}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that 

$P= VI$
$\implies P= V\times \dfrac{V}{R}$
$\implies P = \dfrac{V^2}{R}$
$\implies R= \dfrac{V^2}{P}$..............(1)
Therefore the option D is wrong .

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

Calculate the energy transferred by a $5  A$ current flowing through a resistor of $20   \ ohms$ for $30$ minutes.

  1. $25 kwh$
  2. $50 kwh$
  3. $2.5 \times {10}^{-2} kwh$
  4. $5 \times {10}^{-2} kwh$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$E = {I}^{2}Rt$
$= {5}^{2} \times 2 \times \dfrac{1}{2}$
$= 2.5 \times {10}^{-2}  kwh$

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

If 2.2 kW power transmits 22000 V in a line of $10 \Omega$ resistance, the value of power loss will be :-

  1. 0.1 W

  2. 10 W

  3. 100 W

  4. 1000 W
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power loss in a line is calculated using the formula P = I^2 * R. First, find current I = P / V = 2200 W / 22000 V = 0.1 A. Then, P_loss = (0.1)^2 * 10 = 0.01 * 10 = 0.1 W.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

An electrician not aware of the colour coding of resistors connected two resistors A and B in series to a 6V battery of internal resistance $3\Omega$ and an ammeter. The ammeter connected in the circuit was not working and hence he disconnected the ammeter from the circuit. The sequence of the colour bands on resistor. A is yellow, violet and brown while that on resistor B is red, violet and black respectively. By using the colour coding of resistors, help the electrician to determine the current flowing through the circuit. 

  1. $12$ mA
  2. $24$ mA
  3. $48$ mA
  4. $32$ mA
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: resistors A and B connected in series to a $6V$ battery of $3\Omega$ internal resistance. The sequence of the colour bands on resistor. A is yellow, violet and brown while that on resistor B is red, violet and black respectively. 

To find the current flowing through the circuit.
Solution:
According to the question the A and B are 3-band resistors, so the thirdcode will be a multiplier.
And using the standard resistor color code table the value of:
yellow is 4, violet is 7, brown is $\times 10^1$, red is 2, black is $\times 10^0$.
The corresponding value of A and B are:
A= $47\times 10^1=470\Omega$, B= $27\times 10^0=27\Omega$
Now A and B are in series are also the internal resistance will also be in series. So the effective resistance of the circuit will be $R _{eq}=470+27+3=500\Omega$.
We know according to Ohm's Law,
$V=IR\implies I=\dfrac VR\\implies I=\dfrac 6{500}=0.012A$
Or the current flowing through the circuit = $12mA$

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

The colour code of a resistor is brown, black and brown. Then the value of resistance is _____

  1. $10$ $\Omega$
  2. $100\ m \Omega$
  3. $0.1\ k$$\Omega$
  4. $1000$ $\Omega$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

First two colours says about significant digits of resistance and the third colour gives multiplier.


Brown colour gives digit $1$, brown colour gives digit $0$ and for multiplier brown colour gives $10^1$.

Hence, resistance  $=R=10\times 10^1=100\Omega$

$\implies R=0.1k\Omega$

Answer-(C)

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

The resistance of hot turngsten filament is about $10$ times the cold resistance. What will  be the resistance of $100W$ and $200V$ lamp when not in use

  1. $40\Omega $
  2. $20\Omega $
  3. $400\Omega $
  4. $200\Omega $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$P= 100W$
$V= 200V$
${ R } _{hot}= 10 \cdot { R } _{ cold }$
We know,
$ P= I \cdot V$
$ \rightarrow P= \cfrac{ V }{ R } \cdot { V }= { V }^{ 2 } \cdot R$
$ \therefore R= \cfrac{{ V }^{ 2 }}{P}$
where, 
$ P= $ Power
$ I = $ Current
$ R = $ Resistance
$ V =$ Voltage
$ { R } _{ hot }= $ Hot Resistance
${ R } _{ cold }=$ Cold Resistance
So, $ { R } _{ hot}= \cfrac {200 \cdot 200}{ 100}= 400 \Omega$
${ R } _{ cold }= \cfrac{{ R } _{ hot }}{ 10 }= \cfrac{ 400 }{ 10 }= 40 \Omega $
$\Rightarrow { R } _{ cold }= 40 \Omega$
$ 40 \Omega $ will be the resistance of $100W$ and $200V$ lamp when not in use.

Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

A 2 m long wire of resistance 4 ohm and diameter 0.64 mm is coated with plastic insulation of thickness 0.06 mm, when a current of 5 ampere flows through the wire, find the temperature difference across the insulation in steady state if $k=0.16\times { 10 }^{ -2 }cal/\left( cm{ C }^{ 0 }sec \right) $ 

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

The temperature difference across cylindrical insulation is given by ΔT = Q × ln(r2/r1) / (2πkL), where Q = I²R. For the given values (wire radius 0.32mm, insulation thickness 0.06mm, k=0.16×10⁻², L=200cm, Q=5²×4=100W), the temperature difference works out to approximately 3°C. The insulation thickness is small compared to wire radius, which affects the logarithmic term in the heat transfer equation.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The current voltage relation of diode is given by $I=\left( { e }^{ 1000V/T }-1 \right)mA$, where the applied voltage $V$ is in kelvin. If a student makes an error measuring $\pm 0.01V$ while measuring the current of $5mA$ at $300k$, what will be the error in the value of current in mA?

  1. $0.2mA$
  2. $0.02mA$
  3. $0.5mA$
  4. $0.05mA$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length
Unaware about the fact that analog ammeters and voltmeters can also have zero error, a student recorded following readings while determine resistance by Ohm's law 
If the ammeter has no zero error, the zero error is the voltmeter is 
 Obs, No. Voltage/V  Current/mA   Obs No.  Voltage/V  Current/mA
 1  1.0  40  4  7.0  160
 2  3.0  80  5  9.0  200
 3  5.0  120      
  1. $-1 V$
  2. $-1.5V$
  3. $0.5V$
  4. $1V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Ohm's law V = IR, the slope of the V-I graph gives the resistance. Plotting the points (1.0, 40), (3.0, 80), (5.0, 120), (7.0, 160), (9.0, 200) shows a linear relationship V = 0.05I - 1.0. When I = 0, V = -1.0V, indicating a zero error of -1V.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

A resistor of $10 k\Omega$ having tolerance 10% is connected in series with another resistor of $20k\Omega$ having tolerance 20%. The tolerance of the combination will be:

  1. 10%

  2. 13%

  3. 30%

  4. 20%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In series effective resistance $=R _S=(10k\Omega \pm 10$%)+$(20k\Omega \pm 20$%)$=(30k\Omega \pm 30$%)
$\therefore$ Tolerance of the combination $=( \pm 30$%)

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

Find the percentage error in specific resistance given by $\displaystyle \rho=\frac{\pi r^{2}R}{l}$ where r is the radius having value $\displaystyle \left ( 0.2\pm 0.02 \right )$ cm, R is the resistance of $\displaystyle \left (60\pm 2 \right )\Omega $ and l is length of $\displaystyle \left ( 150\pm 0.1 \right )$ cm.

  1. 5.85%

  2. 11.7%

  3. 23.4%

  4. 35.1%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Applying logarithm on both sides of the given expression and differentiating, 

we get $\dfrac { \Delta \rho  }{ \rho  } =\pm (2\dfrac { \Delta r }{ r } +\dfrac { \Delta l }{ l } +\dfrac { \Delta R }{ R } )$
Given : $\Delta r$=0.02cm, $\Delta R$=2 ohm, $\Delta l$=0.1cm


Substituting the values in above expression,
$\dfrac { \Delta \rho  }{ \rho  } =\pm (2\dfrac { 0.02 }{ 0.2 } +\dfrac { 0.1 }{ 150 } +\dfrac { 2 }{ 60 } )=0.234=23.4$%