Current Electricity Previous year Questions with Solutions

Current Electricity is a branch of Physics that deals with the study of electric charges in motion. It focuses on how charges flow through conductors, how electrical energy is transferred, and how resistors, capacitors, and other components behave in circuits. Current electricity is fundamental to understanding electrical circuits, devices, and the various applications of electricity.

1.0Important Definitions

1. Current: Rate of flow of charge through cross-section of a conductor is called electric current.

$I=\frac{\Delta Q}{\Delta t}=\frac{ne}{t}=\frac{C}{s}=\text{ Ampere }$

2. Drift Velocity \left(\vec{v}_d\right)- It is the mean velocity acquired by the free electrons of a conductor in the inverse direction of the externally applied electric field.

${\vec{V}}_d=\frac{e\vec{E}\tau }{m}$

3.Current Density

• It is defined as the current flowing per unit cross-sectional area of a conductor.
• It gives us information about magnitude and direction of current passing through an area.

$\vec{J}=\frac{I}{A}$

4.Ohm’s Law

• In Physical quantities like temperature,pressure,volume,length,cross-section or nature of the material kept constant then current through a conductor is directly proportional to potential difference applied across it.This is called Ohm’s Law.

$V=IR$

5.Carbon Colour Code Resistor

 6. Kirchhoff’s First Law

• This law is also known as junction law or current law (KCL).According to this-In an electric circuit, the net sum of the currents gathering at any junction in the circuit is zero.
• Sum of the currents arriving the junction is equal to sum of the currents departing the junction

$\sum i=0$

7.Kirchhoff’s Second Law 

• It is also known as loop rule or voltage law (KVL) and according to it in any closed circuit the algebraic sum of e.m.f. and algebraic sum of potential drops is zero. 

$\sum IR+\sum E=0$

8.Cell:A cell is a device that provides the necessary potential difference to maintain a continuous flow of current in an electric circuit. It consists of two electrodes, typically rods or plates, which are immersed in a chemical solution known as the electrolyte. 

Note-The Current Electricity chapter is crucial for JEE Main, contributing approximately 9 to 10% of the total Physics section weightage. Students can expect around 3 to 4 questions from this topic, totaling about 12 marks. Key concepts in this chapter, such as Ohm's Law and Kirchhoff’s Laws, are commonly tested and play a vital role in the exam.

This study material on current electricity is essential for JEE Main preparation as it covers fundamental concepts like Ohm's Law and Kirchhoff's Laws, which are frequently tested in the exam. By mastering these topics, students can solve a wide range of problems, improving their problem-solving skills and boosting their confidence. It also provides a solid foundation for tackling both conceptual and numerical questions, enhancing overall performance in the exam.

2.0Key Concepts To Remember

Current	$I=\frac{\Delta Q}{\Delta t}=\frac{ne}{t}=\frac{C}{s}=\text{ Ampere }$
Instantaneous current	$\begin{array}{c}i=\underset{\Delta t\to 0}{\lim }\left(\frac{\Delta q}{\Delta t}\right)=\frac{dq}{dt}\\ dq=idt\\ q=\int idt\end{array}$
Drift Velocity	${\vec{V}}_d=\frac{e\vec{E}\tau }{m}$
Relation Between Drift Velocity and Current	$i=\frac{dq}{dt}=\frac{enAdx}{dt}=neA\left(\frac{dx}{dt}\right)=neA{V}_d$
Mobility	$\mu =\frac{V_d}{E}=\frac{1}{E}\frac{eE\tau }{m}\Rightarrow \mu =\frac{e\tau }{m}$
Current density	$\begin{array}{r}\vec{J}=\frac{I}{A}\\ I=JA\mathrm{Cos}\theta \\ I=\vec{J}\cdot \vec{A}\\ I=\int J(dA\mathrm{Cos}\theta )\end{array}$
Relation between Current density and Electric Field	$J=\sigma E$
Ohm’s Law	$V=IR$
Electrical Resistance	$R=\rho \frac{l}{A}$
Specific Resistance or Resistivity(\rho)	$\rho =\frac{m}{n{e}^2\tau }$
Temperature dependence of Resistance and Resistivity	$\begin{array}{c}{\rho }_T={\rho }_o\left[1+\alpha \left(T-T_o\right)\right]\text{ (Resistivity) }\\ R_T=R_0[1+\alpha \Delta T]\text{ (Resistance) }\end{array}$
Resistance in Series	$R=R_1+R_2+R_3$
Resistance In Parallel	$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
Kirchhoff’s First Law	$\begin{array}{rl}& \sum i=0\\ & i_2+i_3+i_4=i_1+i_5\end{array}$
Kirchhoff’s Second Law	$\sum IR+\sum E=0$
Terminal Potential Difference(V)	$V=E-Ir$
When Cell is getting Charged	$V=E+Ir$
Combination of Cells in Series	$i=\frac{E_{eq.}}{r_{eq.}+R}$
When Cells supports each other	$i=\frac{E_1+E_2}{R+\left(r_1+r_2\right)}$ T.P.D of First Cell $=E_1-i{r}_1$ T.P.D of Second Cell $=E_2-i{r}_2$
When Cells are Connected with Opposite Polarity	$\begin{array}{c}{E}_1>E_2\\ i=\frac{E_1-E_2}{R+\left(r_1+r_2\right)}\end{array}$ T.P.D of first cell $=E_1-i{r}_1$ T.P.D of second cell $=E_2+i{r}_2$
Combination of Cells in Parallel	$\begin{array}{c}{E}_{eq.}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}+\dots \dots }{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\dots .}\\ \frac{1}{r_{eq.}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\dots \dots \end{array}$
When cells support each other	$\begin{array}{c}{E}_{eq.}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}\\ r_{eq}=\frac{r_1\cdot r_2}{r_1+r_2}\end{array}$
When cells are connected with opposite polarity	$\begin{array}{c}{E}_1>E_2\\ E_{eq}=\frac{E_1{r}_2-E_2{r}_1}{r_1+r_2}\\ r_{eq}=\frac{r_1{r}_2}{r_1+r_2}\end{array}$
Maximum Power Transfer Theorem	$P_{\mathrm{Max}}=\frac{E^2}{4R}$
Electrical Energy And Power	$\begin{array}{c}H=i^{2}Rt=\frac{V^2}{R}=Vit\\ P=\frac{dw}{dt}=\frac{i^{2}Rdt}{dt}=i^{2}R=\frac{V^2}{R}=Vi\end{array}$
Ammeter	$\begin{array}{c}\text{ Shunt- }S=\left[\frac{i_g}{i-i_g}\right]G\\ \Rightarrow S=\left[\frac{1}{\frac{i}{i_g}-1}\right]G\Rightarrow S=\left[\frac{G}{n-1}\right],\text{ where }n=\frac{i}{i_g}\end{array}$
Voltmeter	$\begin{array}{c}H=\left(\frac{V-V_g}{V_g}\right)G=\left(\frac{V}{V_g}-1\right)G\\ H=(n-1)G\text{ where }n=\frac{V}{V_g}\\ V=\text{ Range of Voltmeter }\\ V_g=\text{ Range of Galvanometer }\end{array}$
Meter Bridge	$RQ=PX\Rightarrow X=R\frac{Q}{P}\text{ or }X=\frac{100-l}{l}R$
End Corrections of Meter Bridge	$\alpha =\frac{R_2{l}_1-R_1{l}_2}{R_1-R_2}\text{ and }\beta =\frac{R_1{l}_1-R_2{l}_2}{R_1-R_2}-100$

3.0JEE Main Past Year Questions with Solutions on Current Electricity

Q-1.A wire of length 10 cm and radius $\sqrt{7}\times 10^{-4}m$ connected across the right gap of a meter bridge. When a resistance of 4.5 \Omega is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is R \times 10-7 \Omega m , then value of R is :

1. 63
2. 70
3. 66
4. 35

Solution:

Ans 3

For Null point,

$\frac{4.5}{60}=\frac{R}{40}AlsoR=\rho \frac{l}{A}=\frac{\rho l}{\pi r^2}\begin{array}{rl}& 4.5\times 40=\rho \times \frac{0.1}{\pi \times 7\times 10^{-8}}\times 60\\ & \rho =66\times 10^{-7}\Omega m\end{array}$

Q-2.Three voltmeters, all having different internal resistances are joined as shown in figure.  When some potential difference is applied across A and B, their readings are V₁, V₂ and V₃­.  Choose the correct option.

1. $V_1=V_2$
2. $V_1\ne V_3-V_2$
3. $V_1+V_2>V_3$
4. $V_1+V_2=V_3$

Solution:

Ans (4) From KVL 

$V_1+V_2-V_3=0\Rightarrow V_1+V_2=V_3$

Q-3.Wheatstone bridge principle is used to measure the specific resistance (S­­₁) of given wire, having length L, radius r.  If X is the resistance of wire, then specific resistance is $S_1=X\left(\frac{\pi r^2}{L}\right)$ If the length of the wire gets doubled then the value of specific resistance will be :

Solution:

$S=\frac{I_{g}G}{\left(I-I_g\right)}=\frac{\left(3\times 10^{-3}\right)10}{8-0.003}=3.75\times 10^{-3}\Omega$

Q-4.The deflection in the moving coil galvanometer falls from 25 divisions to 5 divisions when a shunt of $24\Omega$ is applied. The resistance of galvanometer coil will be :

1. $12\Omega$

1. $96\Omega$
2. $48\Omega$
3. $100\Omega$

Solution: Ans(2)

Let $x=\text{ Current/division }$

 After applying shunt 

 $\begin{array}{rl}& \text{ Now }5x\times G=20x\times 24\\ & G=4\times 24=96\Omega \end{array}$

Q-5.Two cells are connected in opposition as shown. Cell E₁ is of 8 V emf and 2 \Omega internal resistance; the cell E₂ is of 2 V emf and 4 \Omega internal resistance. The terminal potential difference of cell E₂ is:

 Solution: Ans (6)

 $I=\frac{8-2}{2+4}=\frac{6}{6}=1A$

Applying Kirchhoff's Law from C to B 

$\begin{array}{rl}& V_C-2-4\times 1=V_B\\ & V_C-V_B=6V\end{array}$

Q-7. When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the same supply, the energy dissipation rate will become:

1. $\frac{1}{4}W$
2. $\frac{1}{2}W$
3. 2 W
4. 4 W

Solution: Ans(4)

$\begin{array}{rl}& \frac{V^2}{R}=W\dots \dots .\\ & \frac{V^2}{\frac{1}{2}\left(\frac{R}{2}\right)}=W^{\prime }\dots .\end{array}$

From (1) and (2)

$W^{\prime }=4W$

Q-8.The resistance per centimeter of a meter bridge wire is r,  with $X\Omega$ resistance in the left gap. Balancing length from left end is at 40 cm with $25\Omega$ resistance in the right gap. Now the wire is replaced by another wire of 2r resistance per centimeter. The new balancing length for same settings will be at

1. 20 cm
2. 10 cm
3. 80 cm
4. 40 cm

Solution: Ans(4)

$\begin{array}{rl}& \frac{25}{r{l}_1}=\frac{X}{r{l}_2}\dots \dots \\ & \frac{25}{r{l}_1^{\prime }}=\frac{X}{2r{l}_2}\dots .\end{array}$

From (1) and (2)

$i_2^{\prime }=l_2=40\mathrm{cm}$

Q-9. By what percentage will the illumination of the lamp decrease if the current drops by 20%?

1. 46%
2. 26%
3. 36%
4. 56%

Solution:Ans(3)

$\begin{array}{rl}& P=i^{2}R\\ & P_{\text{int }}=I_{\text{int }}^{2}R\\ & P_{\text{final }}={\left(0.8I_{\text{int }}\right)}^{2}R\end{array}$

$\%\text{ Change in power }=\frac{P_{\text{final }}-P_{\text{int }}}{P_{\text{int }}}\times 100=(0.61-1)\times 100=-36\%$

Q-10.In the following circuit, the battery has an emf of 2 V and an internal resistance of \frac{2}{3} \Omega. The power consumption in the entire circuit is ______ W.

Solution:

$\begin{array}{rl}& R_{eq}=\frac{4}{3}\Omega \\ & P=\frac{V^2}{R_{eq}}=\frac{4}{\frac{4}{3}}=3W\end{array}$

Q-11.In a metre-bridge when a resistance in the left gap is 2 and unknown resistance in the right gap, the balance length is found to be 40 cm. On shunting the unknown resistance with 2 , the balance length changes by :

1. 22.5 cm
2. 20 cm
3. 62.5 cm
4. 65 cm

Solution:Ans(1)

First Case:

$\frac{2}{40}=\frac{X}{60}\Rightarrow X=3\Omega$

Second case

$X^{\prime }=\frac{2\times 3}{2+3}=1.2\Omega$

$\begin{array}{rl}& \frac{2}{l}=\frac{1.2}{100-l}\\ & 200-2l=1.2l\\ & l=\frac{200}{3.2}=62.5\mathrm{cm}\end{array}$

Balance length change by 22.5 cm

Q-12.The reading in the ideal voltmeter (V) shown in the given circuit diagram is :

1. 5V
2. 10V
3. 0V
4. 3V

Solution: Ans(3)

$\begin{array}{rl}& i=\frac{E_{eq}}{r_{eq}}=\frac{8\times 5}{8\times 0.2}\Rightarrow 25\mathrm{A}\\ & V=E-iR=5-0.2\times 25=0\mathrm{V}\end{array}$

Q-13.A galvanometer has a resistance of 50 and it allows a maximum current of 5 mA. It can be converted into voltmeter to measure up to 100 V by connecting in series a resistor of resistance

1. $5975\Omega$
2. $20050\Omega$
3. $19950\Omega$
4. $19500\Omega$

Solution: Ans(3)

 $\begin{array}{rl}& R=\frac{V}{I_g}-R_g=\frac{100}{5\times 10^{-3}}-50\\ & R=20000-50=19950\Omega \end{array}$

Q-14.The current in a conductor is expressed as t, where I is in Ampere and t is in second. The amount of electric charge that flows through a section of the conductor during $t=1s\text{ to }t=2s$ is ____________ C.

Solution:Ans (22)

$\begin{array}{rl}& q={\int }_1^{2}idt={\int }_1^{2}idt={\int }_1^2\left(3t^2+4t^3\right)dt\\ & q={\left|t^3+t^4\right|}_1^2\end{array}$

q = 22C

Q-15.Twelve wires each having resistance $2\Omega$ are joined to form a cube. A battery of 6 V emf is joined across point a and c. The voltage difference between e and f is_____ V.

Solution:Ans(1)

 From symmetry,current through e-b and g-d = 0

Q-16.A wire of resistance R and radius r is stretched till its radius becomes r/2. If the new resistance of the stretched wire is x R, then the value of x is _________.

Solution:Ans(16)

$R=\rho \frac{l}{A},R\propto \frac{1}{r^2}$

As we stretch the wire ,its length will increase but its radius will decrease keeping the volume constant.

$\begin{array}{rl}& V_i=V_f\\ & \pi r^{2}l=\pi \frac{r^2}{4}{l}_f\\ & l_f=4l\\ & \frac{R_{\text{new }}}{R_{\text{old }}}=\left(\frac{4l}{\frac{r^2}{4}}\right)\frac{r^2}{l}=16\\ & R_{\text{new }}=16R\\ & x=16\end{array}$

Q-17.A galvanometer has a coil of resistance $200\Omega$ with a full scale deflection at $20\mu$A. The value of resistance to be added to use it as an ammeter of range (0–20) mA is:

Solution:(2)

$\begin{array}{rl}& G=200\Omega ,i_g=20\mu A\\ & i=i_g\left(\frac{G}{S}+1\right)\\ & \Rightarrow 20\times 10^{-3}=20\times 10^{-6}\left(\frac{200}{S}+1\right)\\ & \Rightarrow \frac{200}{S}=999\\ & \Rightarrow S\approx 0.2\Omega \end{array}$

Q-18.The equivalent resistance between A and B is:

1. $18\Omega$
2. $25\Omega$
3. $37\Omega$
4. $19\Omega$

Solution: Ans(4)

 $R_{eq}=6+5+8=19\Omega$

Q-19.At room temperature (27ºC), the resistance of a heating element is $50\Omega$. The temperature coefficient of the material is 2.4 × 10^–4 ºC^–1. The temperature of the element, when its resistance is $62\Omega$ is …………. ºC.

Solution:Ans(1027)

$\begin{array}{rl}& R=R_o(1+\alpha \Delta T)\\ & 62=50\left[1+2.4\times 10^{-4}\Delta T\right]\\ & \Delta T=1000^{\circ }\mathrm{C}\\ & \Rightarrow T-27=1000\Rightarrow T=1027^{\circ }\mathrm{C}\end{array}$

Q-20.An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is :

(1) 12.5 W                     (2) 25 W         (3) 50 W                  (4)100 W

Solution:Ans(1)

 Rated power and voltage gives resistance

$\begin{array}{rl}& R=\frac{V^2}{P}=\frac{(200{)}^2}{50}=\frac{40000}{50}\\ & R=800\\ & P=\frac{{\left(V_{\text{applied }}\right)}^2}{R}=\frac{(100{)}^2}{800}\\ & P=12.5\mathrm{W}\end{array}$

Q-21.The current flowing through the $1\Omega$ resistor is  A. The value of n is  ________.

  Solution:Ans(25)

 $\begin{array}{rl}& \frac{y-5}{2}+\frac{y-0}{2}+\frac{y-x+10}{1}=0\\ & y-5+y+2y-2x+20=0\\ & 4y-2x+15=0\dots \dots \dots (1)\\ & \frac{x-5}{4}+\frac{x-0}{4}+\frac{x-10-y}{1}=0\\ & x-5+x+4x-40-4y=0\\ & 6x-4y-45\dots \dots \dots (2)\\ & x=\frac{15}{2}\\ & 4y-15+15=0\Rightarrow y=0\\ & i=\frac{y-x+10}{1}=\frac{0-7.5+10}{1}=2.5\mathrm{A}\\ & i=2.5\mathrm{A}=\frac{n}{10}\mathrm{A}\\ & n=25\end{array}$