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

If current is flowing through a 10 $\Omega$ resistor, then indicate in which case the maximum heat will be generated?

  1. Current of $5\ A$ flows for $2\ \text{minutes}$.
  2. Current of $4\ A$ flows for $3\ \text{minutes}$.
  3. Current of $3\ A$ flows for $6\ \text{minutes}$.
  4. Current of $2\ A$ flows for $5\ \text{minutes}$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat produced is given by
$Q = I^2 R t$

Option A:
$5\ A$ flows for $2\ \text{minutes}$
$Q = 5^2 \times 10 \times 2 = 500\ J$

Option B:

$4\ A$ flows for $3\ \text{minutes}$

$Q = 4^2 \times 10 \times 3 = 480\ J$


Option C:
$3\ A$ flows for $6\ \text{minutes}$
$Q = 3^2 \times 10 \times 6 = 540\ J$

Option D:
$2\ A$ flows for $5\ \text{minutes}$
 $Q = 2^2 \times 10 \times 5 = 200\ J$


Hence, option C is correct.

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

Maximum power in a $0.5 \Omega$ resistance connected with two batteries of 2V emf  and $1 \Omega$ internal resistance in parallel is 

  1. $2 W$
  2. $1.28 W$
  3. $\dfrac{8}{9} W$
  4. $3.2 W$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two identical batteries of emf E = 2V and internal resistance r = 1 ohm in parallel have an equivalent emf of 2V and internal resistance r_eq = 1/2 = 0.5 ohm. The maximum power is transferred to the load resistance R when R equals the internal resistance of the source, R = r_eq = 0.5 ohm. The maximum power formula is P = E^2 / (4 * r_eq), which evaluates to 2^2 / (4 * 0.5) = 4 / 2 = 2 W.

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

How the small bulbs, which are used for decoration purpose,are connected?

  1. in parallel

  2. in series

  3. in mixed order

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

Bulbs used for decoration purposes are connected end to end, that is, in series. This ensures that same amount of current flows through them and the voltage divides across the bulbs as required for low voltage operation. Thus when one bulb goes out, due to being in series, all the bulbs stop glowing.

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

Electric current in a conductor causes heat by:

  1. creating an open circuit

  2. decreasing the watts used

  3. increasing the resistance in the line

  4. decreasing the resistance in the line
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Joule's Law, heat is generated when current flows through a conductor due to resistance. Higher resistance leads to more heat production.

Multiple choice

How does the cold setting work on a blow dryer?

  1. stops all flow of current to the heating element and to the blower

  2. increases the flow of current to the blower and to the heating element

  3. stops all flow to the heating element but still allows flow to the blower

  4. increases flow to heating element and still flows current to the blower
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The cold setting on a blow dryer works by disconnecting the heating element from the circuit while keeping the fan (blower) running.