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JEE 2022 | Class 12
JEE

JEE Important Chapter: Current Electricity

Current Electricity Important Concepts for JEE

Current Electricity Important Concepts for JEE

The chapter Current Electricity is a continuation of the concepts related to the electric current covered in the previous classes and consists of s 


Important Topics of Current Electricity

  • Electric current density and drift velocity

  • Ohm’s law and temperature depend on resistivity.

  • Grouping of Resistors

  • Colour code of carbon resistor

  • Grouping of cells

  • Kirchhoff’s Law

  • Wheatstone bridge

  • Metre bridge

  • Potentiometer 


Important Concepts of Current Electricity

Concepts

Key Points of the Concepts

  1. Electric current density and drift velocity

  • Electric current is the rate of flow of charge per unit time given by the current electricity formula,

$\lim_{\Delta t \rightarrow 0}\dfrac {\Delta q}{\Delta t}$

  • Electric current density (J) is a vector quantity  and is equal to electric current per unit area.

$J=\dfrac{I}{A}$

  • Drift velocity is the velocity acquired by the electron due to an applied electric field. The relation between current density (J) and drift velocity is 

$J=nev_d$

Where n is the number of charge carriers per unit volume.

Relaxation time (τ) is the time taken by electrons between two successive collisions.

  1. Ohm’s law and temperature depend on resistivity.

  • Ohm’s law states that the current passing through a conductor is directly proportional to the potential difference across it at a constant temperature. The formula for resistance is given by ohm’s law.

$R=\dfrac{V}{I}$

  • Resistivity (𝜌) is the reciprocal of conductivity (𝜎) and both depend on the materials and temperature only. The unit of resistivity is Ωm and the unit of conductivity is S/m (siemen per metre)

$\rho=\dfrac{1}{\sigma}$ 

Resistivity increases with temperature.

  1. Grouping of Resistors

  • For resistors in series connection, the current through the resistors is the same.

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$R_{eq}=R_1+R_2+R_3$


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  • For resistors in parallel combination, the potential difference across each resistor is the same.

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$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$


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  1. Colour code of carbon resistor

  • The resistance value of carbon resistors is marked on it using colour bands. Each colour denotes a digit from 0 to 9. The first two colour bands represent the first two digits and the third band represents the multiplier of 10. The fourth band represents the tolerance level. The numerical digit of each colour is given below

  • Black - 0
  • Brown - 1
  • Red - 2
  • Orange - 3
  • Yellow - 4
  • Green - 5
  • Blue - 6
  • Violet - 7
  • Grey - 8
  • White - 9
  • Gold - 5% tolerance
  • Silver - 10 %



  1. Grouping of cells

  • Internal resistance(r) is the resistance offered within the cell or battery against the current.

  • When cells are connected in a series combination in such a way that the positive terminal is connected to the negative terminal of another cell, the net emf of the combination of cells is given by 

$\varepsilon_{eq}=\varepsilon_1+\varepsilon_2+\varepsilon_3$

The effective internal resistance has to be calculated based on the formula of equivalent resistance in series combination.

$r_{eq}=r_1+r_2+r_3$



  • When cells are connected in parallel combination, the effective emf of the combination of cells and effective internal resistance is given by the formula

$\dfrac{\varepsilon_{eq}}{r_{eq}}=\dfrac{\varepsilon_1}{r_1}+\dfrac{\varepsilon_2}{r_2}+\dfrac{\varepsilon_3}{r_3}$

$\dfrac{1}{r_{eq}}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}$


  1. Kirchhoff’s Law

  • Kirchhoff’s current law is based on the law of conservation of charge. According to the law, the current entering and leaving a junction are equal

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$I_1+I_2+I_3=I_4+I_5$

  • Kirchhoff’s voltage law is based on the law of conservation of energy. It states that the algebraic sum of voltage within a loop is always equal to zero.

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$V_{AB}+V_{BC}+V_{CD}++V_{DA}=0$

  1. Wheatstone bridge

  • A galvanometer is connected across B and D and a battery is connected across A and C of a Wheatstone bridge.


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  • A Wheatstone bridge is said to be balanced if no current passes through the galvanometer and the condition for the Wheatstone bridge to get balanced is given by 

$\dfrac{R_2}{R_{1}}=\dfrac{R_4}{R_3}$

  1. Metre bridge

  • Metre bridge uses the principle of the Wheatstone bridge to calculate the unknown resistance.

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  • By knowing the balance length (l) when the current through galvanometer is used, we can find the value of the unknown resistance using the formula given by

$\dfrac{R}{S}=\dfrac{l}{100-l}$

  1. Potentiometer 

  • Potentiometer is a device that measures the potential difference across two points or emf of a cell without drawing current from it. The potentiometer is more accurate than a voltmeter.

  • The potentiometer is used to compare the emf of two cells and the circuit diagram is given below.


The balancing length (l1) for the cell having emf E1 is obtained when switch s1 is only closed and the balancing length (l2) for the cell having emf E2 is obtained when switch s2 is only closed. The ratio of the emfs of the two cells is given by the formula

$\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$

  • The internal resistance of a cell can be calculated using potential. The experimental set-up is given below.



The balancing length (l1) is obtained when the switch S is open and balancing length (l2) is obtained when the switch S is closed. The formula for calculating the internal resistance (r) is given by

$r=\left (\dfrac{l_1}{l_2}-1\right )R$


List of Important Formulae

Sl. No

Name of the Concept

Formula


Relation between current density(J) and conductivity(𝜎)

$\vec J=\sigma \vec E$

Where E is the electric field


Relation between resistivity(𝜌) and relaxation time (𝜏)

$\rho=\dfrac{m}{ne^2\tau}$

Where m is the mass of an electron, e is charge of an electron and n is the charge carrier density

  1. .

Resistance(R) and conductance (C)

$R=\dfrac{1}{C}t$


Resistance of conductor having length (l) and area (A)

$R=\rho\dfrac{l}{A}$


Temperature dependance of resistivity

$\rho_2 =\rho_1(1+\alpha\Delta T)$

Where ΔT is the change in temperature and  ⍺ is the temperature coefficient of resistivity


Heat energy(H) when a current (I) is passing through a resistor (R)

$H=I^2Rt$

$H=\dfrac{V^2}{R}t$

$H=IVt$

Where t is the time of flow of current (I).

  1. j

Power delivered (P) when a current (I) is passing through a resistor (R)

$P=I^2R$

$P=\dfrac{V^2}{R}$

$P=IV$


Current through resistors in parallel



$I_1=I\left(\dfrac{R_2}{R_1+R_2}\right)$

$I_2=I\left(\dfrac{R_1}{R_1+R_2}\right)$


Potential difference across resistors in series

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$V_1=V\left(\dfrac{R_1}{R_1+R_2+R_3}\right)$

$V_2=V\left(\dfrac{R_2}{R_1+R_2+R_3}\right)$

$V_3=V\left(\dfrac{R_3}{R_1+R_2+R_3}\right)$


Solved Examples 

1. The internal resistance of a cell of emf 2V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. Find the voltage across the cell.

Sol:



Given:

Emf of the cell, E = 2V

Internal resistance of the cell, r = 0.1 Ω

Resistance connected to the cell, R = 3.9 Ω

Let V be the terminal voltage across the cell.

We know that the voltage drop across the resistor, 

$V=IR$....(1)

Applying Kirchhoff's voltage law,

$E-V-Ir=0$

$E-IR-Ir=0$

$2V-3.9I-0.1I=0$

$I=\dfrac{2}{4}=0.5~A$

Substitute the value of current I in equation (1) to obtain the voltage across the cell V.

$V=0.5\times 3.9=1.95~V$


Therefore, the voltage across the cell is 1.95 V

Key point: Since resistor R and the cell are in parallel connection, the voltage across them are equal.


2. The resistance of a wire is R ohm. If it is melted and stretched to n times its original length, its new resistance will be_________.

Sol:

Given:

The resistance of the wire is R. 

Let the length and area of the cross-section of wire be l and A, respectively and the resistivity of the wire be 𝜌. 

If the length of the wire is stretched to n times, then the area of the cross-section of the wire is reduced by n times.

Let the length and area of the cross-section of the stretched wire be l’ and A’, respectively.

The resistance of the stretched wire is given by

$R'=\rho\dfrac{l'}{A'}$

$R'=\rho\dfrac{(nl)}{\left(\dfrac{A}{n}\right)}$

$R'=\rho\dfrac{n^2l}{A}$

$R'=n^2\times\left(\rho\dfrac{l}{A}\right)$

$R'=n^2R$

Therefore, the resistance of the stretched wire is n2 times the initial resistance.

Key point: When a wire is stretched, its length increases and the area of cross-section decreases while keeping the volume constant.


Previous Years’ Questions from JEE Paper

1. Five equal resistances are connected in a network as shown in the figure. The net resistance between points A and B is______. (Feb -2021)

a. 3R/2

b. R/2

c. R

d. 2R



Sol:

If you observe the diagram closely, you can see that the above circuit diagram is a balanced Wheatstone bridge since all the resistances are the same. Therefore, it satisfies the condition


$\dfrac{R_2}{R_{1}}=\dfrac{R_4}{R_3}$



Since it is balanced, the current does not pass through the resistor in the bridge and can be removed while calculating the effective resistance.



The two 2R resistors are in parallel, so the equivalent resistance between A and B is


$R_{eq}=\dfrac{2R\times 2R}{2R+2R}$

$R_{eq}=\dfrac{4R^2}{4R}$

$R_{eq}=R$


Therefore, the equivalent resistance between A and B is R and the correct answer is option C.

Key point to remember: When a difficult circuit diagram having resistors are given, always look for the possibility of a balanced Wheatstone bridge.


2. The length of a potentiometer wire is 1200 𝑐𝑚 and it carries a current of 60 𝑚𝐴. For a cell of emf 5 𝑉 and internal resistance of 20Ω, the null point on it is found to be at 1000 𝑐𝑚. The resistance of the whole wire is_____. (JEE 2020)

a. 80Ω

b. 100Ω

c. 120Ω

d. 60Ω



Sol: 

Given:

The null point of the potentiometer = 1000 cm

The current through the potentiometer wire, I = 60 mA

Emf of the cell in the secondary circuit, E = 5 V

At the null point, no current passes through the galvanometer and hence no current is drawn from the cell. Therefore, the emf of the cell is equal to the potential difference across the section of the wire having a length of 1000 cm.


So, the resistance of the wire of 1000 cm (R1) can be calculated.

$R_1=\dfrac{E}{I}$

$R_1=\dfrac{5~V}{60~\text{mA}}$

$R_1=83.33~ \Omega $


Let the resistance of the wire per unit length be r.


$r=\dfrac{83.33~ \Omega}{1000~\text{cm} }$

$r=83.33\times 10^{-3}~ \Omega/cm$

Now, the resistance of the 1200 cm potentiometer wire (R) is obtained by

$R=83.33\times 10^{-3}~ \Omega/cm~\times 1200~\text{cm}$

$R=100~\Omega$

Therefore, the correct answer is option B.

Key point to remember: At the balancing point, no current is drawn from the cell and hence the terminal voltage and emf are equal.


Practice Questions

1. A uniform wire of resistance 9 Ω is cut into three equal parts. They are connected in the form of an equilateral  triangle ABC. A cell of emf 2V and negligible internal  resistance is connected across B and C. Potential difference across AB is____. (Ans: 1 V)


2. A filament bulb (500 W, 100V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and consumes 500 W. The value of R is_______. (Ans: 26 Ω)


Conclusion

In this article, we have provided important information regarding the chapter on Current Electricity, such as important concepts, types of current electricity formulas, circuit problems, etc. More focus has to be given to circuit problems and measuring instruments like metre bridge and potentiometer to score high marks for JEE.

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Frequently Asked Questions
FAQ

1. What is the weightage of the current electricity in JEE?

Around 2-3 questions are asked in JEE from various parts of the chapter. Most of the questions are based on electric current example problems and questions can also be asked by linking them to concepts from other chapters.

2. What section has to be given more importance in the chapter on current electricity?

More focus must be laid on solving electric current example problems and getting the answer without committing any mistakes. Therefore, students should be able to solve any problems related to this chapter by practising more and more problems.

3. How to score good marks in JEE from the chapter on Current Electricity?

First of all, learn and understand the concepts and current electricity formulae in each chapter in depth. This should be followed by solving all the problems given in NCERT textbooks. Then go through the question papers of the last 20 years. Try to solve them and observe the pattern to score good marks in JEE.