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JEE Main · Physics

Current Electricity Practice Questions

Generate JEE Main level questions on Current Electricity. Focus on Ohm's law, Kirchhoff's laws, and Wheatstone bridge.

381 questions · 20 PYQs · 0 AI practice · JEE Main 2027

Q121JEE Main 2023NAT

The number density of free electrons in copper is nearly $8 \times 10^{28} {\text{m}}^{-3}$ . A copper wire has its area of cross section $=2 \times 10^{-6} {\text{m}}^2$ and is carrying a current of $3.2 {\text{A}}$ . The drift speed of the electrons is ________ $\times 10^{-6} {\text{ms}}^{-1}$ [8-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{I}}=\;\text{ neAvd }\;$ $\;\Rightarrow 3.2=8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6} ( {\text{v}}_{ {\text{d}}})$ $\;\Rightarrow {\text{v}}_{ {\text{d}}}=\;\frac{1}{8 \times 10^{-6} \times 10^9}$ $\;\Rightarrow {\text{v}}_{ {\text{d}}}=125 \times 10^{-6} {\text{m}} / {\ s}$

Q122JEE Main 2023MCQ

The drift velocity of electrons for a conductor connected in an electrical circuit is ${\text{V}}_{ {\text{d}}}$ . The conductor in now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{V}}_{ {\text{d}}}=\;{ {\text{eE}}}/{ {\text{m}}} \tau$ that is independent of area

Q123JEE Main 2023MCQ

With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is(A) directly proportional to the length of the potentiometer wire(B) directly proportional to the potential gradient of the wire(C) inversely proportional to the potential gradient of the wire(D) inversely proportional to the length of the potentiometer wireChoose the correct option for the above statements: [29-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Sensitivity of potentiometer wire is inversely proportional to potential gradient.

Q124JEE Main 2023NAT

A potential ${\text{V}}_0$ is applied across a uniform wire of resistance ${\text{R}}$ . The power dissipation is ${\text{P}}_1$ . The wire is then cut into two equal halves and a potential of ${\text{V}}_0$ is applied across the length of each half. The total power dissipation across two wires is $P_2$ . The ratio $P_2: P_1$ is ${\sqrt{x}}: 1$ . The value of $x$ is ________ [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{P}}_2=\;{ {\text{v}}_0^2}/{ (\;{ {\text{R}}}/{2}) }+\;{ {\text{v}}_0^2}/{ (\;{ {\text{R}}}/{2}) }$ $\; {\text{P}}_2=4 {\text{P}}_1$ $\;\;{ {\text{P}}_2}/{ {\text{P}}_1}=\;\frac{4}{1}=\;{\sqrt{ {\text{x}}}}/{1}$ $\; {\text{x}}=16$

Explanation Figure 1

Q125JEE Main 2023NAT

A current of $2 {\text{A}}$ through a wire of cross-sectional area $25.0 {\text{mm}}^2$ . The number of free electrons in a cubic meter are $2.0 \times 10^{28}$ . The drift velocity of the electrons is _______ $\times 10^{-6} {\text{ms}}^{-1}$ (given, charge on electron $=1.6 \times 10^{-19} {\text{C}}$ ). [8-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{I}}={neAV_{d }}$ $\; {\text{V}}_{ {\text{d}}}=\;{ {\text{I}}}/{ {\text{neA}}} \Rightarrow V_d=\;\frac{2}{2 \times 10^{28} \times 1.6 \times 10^{-19} \times 25 \times 10^{-6}}$ $\; {\text{V}}_{ {\text{d}}}=25 {\text{m}} / {\ s}$

Q126JEE Main 2023MCQ

Equivalent resistance between the adjacent corners of a regular ${\text{n}}$ -sided polygon of uniform wire of resistance ${\text{R}}$ would be: [1-Feb-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Suppose resistance of each arm is $r$ , then $r=R / n$ $\;R_{e q(A B)}=\;\frac{R_1 R_2}{R_1+R_2}$ $\;\;\frac{r(n-1) r}{r+(n-1) r}$ $\;=\;\frac{r(n-1) r}{n r}$ $\;=\;\frac{n-1}{n} r$ $\;=\;\frac{(n-1) R}{n^2}$

Explanation Figure 1

Q127JEE Main 2023MCQ

A cell of emf $90 {\text{V}}$ is connected across series combination of two resistors each of $100 Ω$ resistance. A voltmeter of resistance $400 Ω$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be : [24-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{R}}_{ {\text{eq}}}=\;\frac{400 \times 100}{500}+100$ $\;=180 Ω$ $\; {\text{i}}=\;\frac{90}{180}=\;\frac{1}{2} {\text{A}}$ $\;\;\text{ Reading }\;=\;\frac{1}{2} \times \;\frac{400}{500} \times 100$ $\;=40 \;\text{ volt }\;$

Explanation Figure 1

Q128JEE Main 2023MCQ

Figure shows a part of an electric circuit. The potentials at points ${\text{a}}, {\text{b}}$ and ${\text{c}}$ are $30 {\text{V}}, 12 {\text{V}}$ and $2 {\text{V}}$ respectively. The current through the $20 Ω$ resistor will be [6-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Let potential of the junction be ${\text{x}}$ volts using junction law ${\text{i}}_{ {\text{i}}}+ {\text{i}}_2+ {\text{i}}_3=0$ or $\;\frac{x-30}{10}+\;\frac{x-12}{20}+\;\frac{x-2}{30}=0$ $\;\;\text{ or }\; \;\frac{1}{60}[6 x-180+3 x-36+2 x-4]=0$ $\;\;\text{ or }\; \;\frac{1}{60}[11 x-220]=0$ $\;\;\text{ or }\; x=\;\frac{220}{11}=20 {\text{V}}$ current through $20 Ω$ is $=\;\frac{x-12}{20}$ ${\text{i}}_2=\;\frac{20-12}{20}=0.4 {\text{A}}$

Q129JEE Main 2023MCQ

Given below are two statements :Statement I: The equivalent resistance of resistors in a series combination is smaller than least resistance used in the combination.Statement II: The resistivity of the material is independent of temperature.In the light of the above statements, choose the correct answer from the options given below : [15-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

In series, $\; {\text{R}}_{\;\text{eq }\;}= {\text{R}}_1+ {\text{R}}_2+ {\text{R}}_3+. . . .$ $\; {\text{R}}_{\;\text{eq }\;}> {\text{R}}_{\;\text{Greatest }\;}$ Hence, statement-I is false.Resistivity of conductor increases with temperature.Hence, statement-II is also false

Q130JEE Main 2023MCQ

For designing a voltmeter of range $50 {\text{V}}$ and an ammeter of range $10 {\text{mA}}$ using a galvanometer which has a coil of resistance $54 Ω$ showing a full scale deflection for $1 {\text{mA}}$ as in figure.(A) for voltmeter ${\text{R}} \approx 50 {\text{k}} Ω$ (B) for ammeter ${\text{r}} \approx 0.2 Ω$ (C) for ammeter ${\text{r}} \approx 6 Ω$ (D) for voltmeter ${\text{R}} \approx 5 {\text{k}} Ω$ (E) for voltmeter ${\text{R}} \approx 500 Ω$ Choose the correct answer from the options given below : [15-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

For voltmeter, $\; {\text{I}}=\;{50}/{ {\text{R}}+54}=0.001 {\text{A}}$ $\; {\text{R}}=50 {\text{k}} Ω$ For Ammeter, $\; {\text{I}}_{ {\text{r}}}=10-1=9 {\text{mA}}$ $\; {\text{V}}_{ {\text{G}}}= {\text{V}}_{ {\text{r}}}$ $\;1 {\text{mA}} \times 54=9 {\text{mA}} \times {\text{r}}$ $\; {\text{r}}=6 Ω$

Q131JEE Main 2023MCQ

The ${\text{H}}$ amount of thermal energy is developed by a resistor in $10 {\ s}$ when a current of $4 {\text{A}}$ is passed through it. If the current is increased to $16 {\text{A}}$ , the thermal energy developed by the resistor in $10 {\ s}$ will be: [31-Jan-2023 Shift 2]

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📖 Explanation

$H \propto i^2$ for $t=$ constant $\Rightarrow \;\frac{H}{H^{'}}= (\;\frac{4}{16}) ^2$ $\Rightarrow H^{'}=16 H$

Q132JEE Main 2023NAT

In the following circuit, the magnitude of current ${\text{I}}_1$ , is ______ A. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Junction law at A, $\;\frac{x-(y+5)}{1}+\;\frac{x-2}{2}+\;\frac{x-0}{2}=0$ . . . (1)Junction law at B, $\;\frac{y+5-x}{1}+\;\frac{y-0}{1}+\;\frac{y-2}{1}=0$ . . . (2)On solving equation (1) and Equation (2) ${\text{x}}=3$ & ${\text{y}}=0$ At ${\text{D}}$ junction $I_1\;=i_1+i_2$ $I_1\;=\;\frac{y-0}{1}+\;\frac{x-0}{2}$ $\;=\;\frac{0-0}{1}+\;\frac{3-0}{2}$ $I_1\;=1.5 {\text{A}}$

Explanation Figure 2

Q133JEE Main 2023NAT

In a metre bridge experiment the balance point in obtained if the gaps are closed by $2 Ω$ and $3 Ω$ . A shunt of $X Ω$ is added to $3 Ω$ resistor to shift the balancing point by $22.5 {\text{cm}}$ . The value of $X$ is _______ [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\frac{2}{ (\;\frac{3 x}{3+x}) }=\;\frac{40+22.5}{60-22.5}=\;\frac{62.5}{37.5}=\;\frac{5}{3}$ $\;\;\frac{6}{5}=\;\frac{3 x}{3+x}$ $\;6+2 x=5 x \Rightarrow x=2$

Q134JEE Main 2023MCQ

The charge flowing in a conductor changes with time as ${\text{Q}}( {\text{t}})=\alpha {\text{t}}-\beta {\text{t}}^{2+}\gamma {\text{t}}^3$ . Where $\alpha , \beta$ and $\gamma$ are constants. Minimum value of current is : [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;Q= (\alpha t-\beta t^{2+}\gamma t^3)$ $\;i=\;\frac{d Q}{d t}= (\alpha -2 \beta t+3 \gamma t^2)$ $\;\;\frac{d i}{d t}=(3 \gamma t-2 \beta )=0$ $\;\Rightarrow t=\;\frac{\beta }{3 \gamma }$ $\;i= (\alpha -2 \beta t+3 \gamma t^2) = (\alpha -\;\frac{\beta ^2}{3 \gamma })$

Q135JEE Main 2023NAT

When a resistance of $5 Ω$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $250 {\text{mA}}$ , however when $1050 Ω$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ________ $Ω$ . [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;I_g(G)= (250-I_g) 5$ $\;I_g=\;\frac{1250}{5+G} m A$ For voltmeter, $\; {\text{V}}= {\text{I}}_{ {\text{g}}} {\text{R}}$ $\;25= {\text{I}}_{ {\text{g}}}( {\text{G}}+1050)$ From equation (1), $\;25=\;{1250 \times 10^{-3}}/{ {\text{G}}+5}( {\text{G}}+1050)$ $\;20( {\text{G}}+5)= {\text{G}}+1050$ $\;19 {\text{G}}=1050-100$ $\; {\text{G}}=\;\frac{950}{19}=50 Ω$

Explanation Figure 1

Q136JEE Main 2023NAT

In the given circuit the value of $|\;\frac{I_1+I_3}{I_2}|$ is: [1-Feb-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$; {\text{I}}_1= {\text{I}}_2=\;\frac{20-10}{10}=1 {\text{A}}$ ${\text{I}}_3=1 {\text{A}}$ $|\;{ {\text{I}}_1+ {\text{I}}_3}/{ {\text{I}}_2}|=2$

Explanation Figure 2

Q137JEE Main 2023MCQ

A uniform metallic wire carries a current $2 {\text{A}}$ . when $3.4 {\text{V}}$ battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3} {\text{kg}}$ . density is $8.92 \times 10^3 {\text{kg}} / {\text{m}}^3$ and resistivity is $1.7 \times 10^{-8} Ω- {\text{m}}$ . The length of wire is : [25-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{I}}=2 {\text{A}}$ $\;\Delta {\text{V}}=3.4 {\text{V}}$ Using Ohm's Law $\; {\text{R}}=\;\frac{3.4}{2}=1.7 Ω$ $\;1.7=\;{\rho {\text{L}}}/{ {\text{A}}}$ $\; {\text{L}}=\;{1.7( {\text{A}})}/{\rho }$ $\; {\text{M}}=(\;\text{ density volume }\;)$ $\;\;\text{ Volume }\;=\;\frac{8.92 \times 10^{-3}}{8.92 \times 10^3}=10^{-6}$ $\; {\text{L}}^2=\;\frac{1.7}{\rho } (10^{-6}) =\;\frac{1.7}{1.7} \times 10^2$ $\; {\text{L}}=10 {\text{m}}$

Q138JEE Main 2023NAT

The current flowing through a conductor connected across a source is $2 {\text{A}}$ and $1.2 {\text{A}}$ at $0^{\circ} {\text{C}}$ and $100^{\circ} {\text{C}}$ respectively. The current flowing through the conductor at $50^{\circ} {\text{C}}$ will be ______ $\times 10^2 {\text{mA}}$ [12-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{i}}_0 {\text{R}}_0= {\text{i}}_{100} {\text{R}}_{100} \;\; \;\text{ (For same source) }\;$ $\;\Rightarrow 2 {\text{R}}_0=1.2 {\text{R}}(1+100 \alpha )$ $\;\Rightarrow 1+100 \alpha =\;\frac{5}{3} \Rightarrow 100 \alpha =\;\frac{2}{3}$ $\;\Rightarrow 50 \alpha =\;\frac{1}{3}$ $\; \therefore {\text{i}}_{50} {\text{R}}_{50}= {\text{i}}_0 {\text{R}}_0$ $\;\Rightarrow {\text{i}}_{50}=\;{ {\text{i}}_0 {\text{R}}_0}/{ {\text{R}}_{50}}=\;{2 \times {\text{R}}_0}/{ {\text{R}}_0(1+50 \alpha )}=\;\frac{2}{1+\;\frac{1}{3}}=1.5 {\text{A}}$ $\;=15 \times 10^2 {\text{mA}}$

Q139JEE Main 2023MCQ

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 \%$ . The percentage change in voltage sensitivity of the galvanometer will be: [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Current sensitivity $=$ Voltage sensitivity $\times R$ Current sensitivity is made $1.5$ times. $R$ also increase $1.5$ times.Hence voltage sensitivity $=\;\frac{1.5 \times \;\text{ current sensitivity }\;}{1.5 \times R}$ $=$ no change

Q140JEE Main 2023MCQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.Assertion A : For measuring the potential difference across a resistance of $600 Ω$ , the voltmeter with resistance $1000 Ω$ will be preferred over voltmeter with resistance $4000 Ω$ .Reason R : Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.In the light of the above statements, choose the most appropriate answer from the options given below. [1-Feb-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Error of voltmeter decreases with increase in its resistance.

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