Current Electricity – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions Download eSaral app  for free study material and video tutorials.SimulatorPrevious Years JEE Advanced Questions
Q. For the circuit shown in the figure –(A) the current I through the battery is 7.5 mA(B) the potential difference across $\mathrm{R}_{\mathrm{L}}$ is 18V(C) ratio of powers dissipated in $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ is 3(D) if $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ are interchanged, magnitude of the power dissipated in $\mathrm{R}_{\mathrm{L}}$ will decrease by a factor of 9 [IIT-JEE 2009]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,D)

Q. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistance $\mathrm{R}_{100} \cdot \mathrm{R}_{60} \mathrm{and} \mathrm{R}_{40}$, respectively, the relation between these resistances is $(\mathrm{A}) \frac{1}{\mathrm{R}_{100}}=\frac{1}{\mathrm{R}_{40}}+\frac{1}{\mathrm{R}_{60}}$$(\mathrm{B}) \mathrm{R}_{100}=\mathrm{R}_{40}+\mathrm{R}_{60}(C) \mathrm{R}_{100}>\mathrm{R}_{60}>\mathrm{R}_{40}$$(\mathrm{D}) \frac{1}{\mathrm{R}_{100}}>\frac{1}{\mathrm{R}_{60}}>\frac{1}{\mathrm{R}_{40}}$ [IIT-JEE 2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (D)

Q. To verify Ohm’s law, a student is provided with a test resistor $\mathrm{R}_{\mathrm{T}}$, a high resistance $R_{1}$, a small resistance $\mathrm{R}_{2}$, two identical galvanometers $\mathrm{G}_{1}$ and $\mathrm{G}_{2}$, and a variable voltage source V. The correct circuit to carry out the experiment is :– [IIT-JEE 2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C)

Q. Consider a thin square sheet of side L and thickness t, made of a material of resistivity . The resistance between two opposite faces, shown by the shaded areas in the figure is –(A) directly proportional to L(B) directly proportional to t(C) independent of L(D) independent of t [IIT-JEE 2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C)$\mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{A}}=\frac{\rho \mathrm{L}}{\mathrm{Lt}}=\frac{\rho}{\mathrm{t}} \Rightarrow$ independent of $\mathrm{L}$

Q. When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor R, the rate of heat produced in R is $\mathbf{J}_{1}$. When the same batteries are connected in parallel across R, the rate is $\mathrm{J}_{2} \cdot$ If $\mathrm{J}_{1}=2.25 \mathrm{J}_{2}$ then the value of R in $\Omega$ is – [IIT-JEE 2010]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. 4$J_{1}=\left(\frac{2 \varepsilon}{R+2}\right)^{2} R$ and $J_{2}=\left(\frac{\varepsilon}{R+1 / 2}\right)^{2} R$ as $\frac{J_{1}}{J_{2}}=2.25$ so $\frac{4 \varepsilon^{2}}{(R+2)}=2.25 \frac{4 \varepsilon^{2}}{(1+2 R)^{2}} \Rightarrow R=4 \Omega$

Q. Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is –[IIT-JEE 2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. 5$E_{A B}=\frac{\frac{E_{1}}{R_{1}}+\frac{E_{2}}{R_{2}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}}=\frac{\frac{6}{1}+\frac{3}{2}}{\frac{1}{1}+\frac{1}{2}}=\frac{\frac{15}{2}}{\frac{3}{2}}=5$ volt

Q. A meter bridge is set-up as shown, to determine an unknown resistance ‘X’ using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of ‘X’ is-(A) 10.2 ohm (B) 10.6 ohm (C) 10.8 ohm (D) 11.1 ohm[IIT-JEE 2011]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B)Apply condition of wheatstone bridge, $\frac{x}{52+1}=\frac{10}{48+2} \Rightarrow x=\frac{10}{50} \times 53 \Rightarrow \mathrm{x}=10.6 \Omega$

Q. For the resistance network shown in the figure, choose the correct option(s).(A) the current through PQ is zero(B) $\mathrm{I}_{1}=3 \mathrm{A}$(C) The potential at S is less than that at Q(D) $\mathrm{I}_{2}=2 \mathrm{A}$ [IIT-JEE 2012]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A,B,C,D)Since resitances in two arms upper and lower are in same ratio.Current through PQ is zero. $I_{2}=\frac{12}{6}=2 A$, $I_{1}=\frac{12}{6}+\frac{12}{12}=3 A$ and Potential of $\mathrm{S}$ is less than that at $\mathrm{Q}$

Q. Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K ?(A) 4 if wires are in parallel(B) 2 if wires are in series(C) 1 if wires are in series(D) 0.5 if wires are in parallel [JEE Advanced 2014]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B,D)Resistance of heater $1, \mathrm{R}=\frac{4 \rho \mathrm{L}}{\pi \mathrm{d}^{2}}$Resistance of heater $2, R_{1}=\frac{R}{4}, R_{2}=\frac{R}{4}$Series $\mathrm{R}_{\mathrm{net}}=\frac{\mathrm{R}}{2}$Power $=2 \frac{\mathrm{V}^{2}}{\mathrm{R}}$$\Rightarrow \quad power is twice, hence time is \frac{1}{2}time =\frac{1}{2} of 4 \min =2 \minParallel \mathrm{R}_{\mathrm{Net}}=\frac{\mathrm{R}}{8}Power = 8 timestime =\frac{\mathrm{R}}{8} times =\frac{\mathrm{R}}{8} \times 4 \min =0.5 \mathrm{sec} Q. Two ideal batteries of emf \mathrm{V}_{1} and \mathrm{V}_{2} and three resistances \mathrm{R}_{1}, \mathrm{R}_{2} and \mathrm{R}_{3} are connected as shown in the figure. The current in resistance \mathrm{R}_{2} would be zero if :-(A) \mathrm{V}_{1}=\mathrm{V}_{2} and \mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}(B) \mathrm{V}_{1}=\mathrm{V}_{2} and \mathrm{R}_{1}=2 \mathrm{R}_{2}=\mathrm{R}_{3}(C) \mathrm{V}_{1}=2 \mathrm{V}_{2} and 2 \mathrm{R}_{1}=2 \mathrm{R}_{2}=\mathrm{R}_{3}(D) 2 \mathrm{V}_{1}=\mathrm{V}_{2} and 2 \mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3} [JEE Advanced 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,B,D)since current through \mathrm{R}_{2} is zeroHence \left[\frac{\mathrm{V}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{R}_{3}}\right]The above equation is satisfied by options (A, B, D) Q. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 \Omega resistance, it can be converted into a voltmeter of range 0 – 30 V. If connected to a \frac{2 \mathrm{n}}{249} \Omega resistance, it becomes an ammeter of range 0 – 1.5 A. The value of n is :- [JEE Advanced 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (5) Q. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 \Omega, as shown in the figure. The least count of the scale used in the metre bridge is 1mm. The unknown resistance is :-(A) 60 \pm 0.15 \Omega(B) 135 \pm 0.56 \Omega(C) 60 \pm 0.25 \Omega(D) 135 \pm 0.23 \Omega [JEE Advanced 2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C)For meter bridge,\mathrm{R}=\mathrm{X}\left(\frac{\ell}{100-\ell}\right)$$=90\left(\frac{\ell}{100-\ell}\right)$$=90\left(\frac{40}{60}\right)v=60 \Omegaerror in R : (erro = least count)\frac{\mathrm{d} \mathrm{R}}{\mathrm{R}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}+\frac{\mathrm{d} \ell}{\ell}+\frac{\mathrm{d}(100-\ell)}{(100-\ell)}$$=0+\frac{\operatorname{lmm}}{40 \mathrm{cm}}+\frac{1 \mathrm{mm}}{60 \mathrm{cm}}$

Q. An infinite line charge of uniform electric charge density $\lambda$ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity $\varepsilon$ and electrical conductivity $\sigma$. The electrical conduction in the material follows Ohm’s law. Which one of the following graphs best describes thesubsequent variation of the magnitude of current density j(t) at any point in the material? [JEE Advanced 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)This is the problem of RC circuit where the product RC is a constant.So due to leakage current, charge & current density will exponentially decay & will become zero at infinite time. So correct answer is (A)for any small elementResistance $\mathrm{R}=\frac{\mathrm{dr}}{\sigma(2 \pi r \ell)}$Capacitance $\mathrm{C}=\frac{\in 2 \pi \mathrm{r} \ell}{\mathrm{dr}}$Product $\mathrm{R} \times \mathrm{C}=\frac{\epsilon}{\sigma}=\mathrm{constant}$$\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\left(\frac{\mathrm{t} \sigma}{\epsilon}\right)}$$\mathrm{I}=\frac{\mathrm{d} q}{\mathrm{dt}}=\frac{\mathrm{q}_{0} \sigma}{\epsilon} \mathrm{e}^{-\left(\frac{\mathrm{t} \sigma}{\epsilon}\right)}$

Q. An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?(A) The temperature distribution over the filament is uniform(B) The resistance over small sections of the filament decreases with time(C) The filament emits more light at higher band of frequencies before it breaks up(D) The filament consumes less electrical power towards the end of the life of the bulb [JEE Advanced 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (C,D)Because of non-uniform evaporation at different section, area of cross-section would be different at different sections.Region of highest evaporation rate would have rapidly reduced area and would become break up cross-section.Resistance of the wire as whole increases with time.Overall resistance increases hence power decreases. At break up junction temperature would be highest, thus light of highest band frequency would be emitted at those cross-section.

Q. Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers $\mathrm{R}_{\mathrm{c}}<\mathrm{R} / 2$, which of the following statement(s) about any one of the galvanometers is(are) true(A) The maximum voltage range is obtained when all the components are connected in series(B) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer(C) The maximum current range is obtained when all the components are connected in parallel(D) The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors. [JEE Advanced 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B,C)Range $=\mathrm{i}_{\mathrm{g}}\left(2 \mathrm{R}_{\mathrm{C}}+2 \mathrm{R}\right)$Range $=\mathrm{i}_{\mathrm{g}}+4 \mathrm{i}_{\mathrm{g}} \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}}$$=2 \mathrm{i}_{\mathrm{g}}\left[\frac{1}{2}+\frac{2 \mathrm{R}_{\mathrm{c}}}{\mathrm{R}}\right]$$=2 \mathrm{i}_{\mathrm{g}}\left(\frac{1}{2}+\frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}}+\frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}}\right)$So (C)

Paragraph 2Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r <<h. Now a high voltage source (HV) is connectedacross the conducting plates such that the bottom plate is at $+V_{0}$ and the top plate at$-\mathbf{V}_{0}$. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collision between the balls and the interaction between them is negligible.(Ignore gravity)
Q. Which of the following statements is correct ?(A) The balls will bounce back to the bottom plate carrying the opposite charge they went up with(B) the balls will execute simple harmonic motion between the two plates(C) The balls will bounce back to the bottom plate carrying the same charge they went up with(D) The balls will stick to the top plate and remain there [JEE Advanced 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (A)Balls placed on +ve plate become positive charge and move upward due to electric field.These balls on colliding with negative plate become negatively charged and move opposite to the direction of electric field.

Q. The average current in the steady state registered by the ammeter in the circuit will be :(A) Proportional to $\mathrm{V}_{0}^{1 / 2}$(B) Proportional to $V_{0}^{2}$(C) Proportional to the potential $\mathrm{V}_{0}$(D) Zero [JEE Advanced 2016]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (B)

• June 30, 2022 at 10:29 pm

Really useful… Thank you esaral.

• June 21, 2022 at 3:36 pm

Hey!

• September 29, 2021 at 1:16 am

Thanku very much 👍

• July 28, 2021 at 5:46 pm

Hmm thats really good question , I like it

• July 2, 2021 at 10:10 pm

really love your clearly explained solutions <3

• July 2, 2021 at 8:31 pm

Like AIM 99999999 🤘🏻

• July 2, 2021 at 8:39 pm

But this site is really useful and made me to know the answers well

• July 2, 2021 at 10:11 pm

100th like was mine 🙂

• June 2, 2021 at 2:50 am

Very usefull!!

• June 2, 2021 at 2:49 am

Very usefull!!

• May 12, 2021 at 4:40 pm

Good question

• May 7, 2021 at 12:04 pm

• September 24, 2020 at 4:34 pm

• August 17, 2022 at 8:38 am

Bakwas

• September 22, 2020 at 4:49 pm

We need after 2016 questions but really questions and there explanations are good

• October 25, 2020 at 11:37 pm

yes please post the questions after 2016 till 2020

• September 21, 2020 at 11:46 am

Thank you

• June 13, 2022 at 7:46 pm

poda pongal

• September 17, 2020 at 9:01 pm

Thanks

• September 13, 2020 at 9:22 pm

Excellent but we need after 2016 also

• August 22, 2020 at 12:36 pm

2017 2018 2019 2020 2021 ???????

• March 18, 2021 at 2:47 pm

unko daale

• August 18, 2020 at 10:40 pm

fan of esaral questions of jee are very well mannered arranged

• August 16, 2020 at 8:49 pm

Ur que are awesome but plz upload some more questions
Thanks !

• July 29, 2020 at 9:03 am

1

• July 11, 2020 at 10:53 pm

• June 17, 2020 at 12:27 am

in the identical galvanometer and identical resistance question wherein diff arrangements of the four elements of the circuit are given, shouldnt the answer be A,C and not B,C?

• September 26, 2021 at 11:39 am

nani?!

• June 16, 2020 at 11:14 pm