PracticeMock TestDashboard Review Mistakes Bookmarks

Mock

Stats Review Mistakes Saved

Home›Physics›Current Electricity

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

🎯 These are sample questions

Get the full Current Electricity experience — 100% free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Q1JEE Main 2026MCQ

For the two cells having same EMF $E$ and internal resistance $r$ , the current passing through the external resistor $6 Ω$ is same when both the cells are connected either in parallel or in series. The value of internal resistance $r$ is $\_\_\_\_$ $Ω$ .

🚀 Solve in Practice Mode

Q2JEE Main 2026MCQ

A wire of uniform resistance $\lambda Ω / m$ is bent into a circle of radius $r$ and another piece of wire with length $2 r$ is connected between points $A$ and $B( AOB )$ as shown in figure. The equivalent resistance between points $A$ and $B$ is $\_\_\_\_$ $Ω$ .

🚀 Solve in Practice Mode

Q3JEE Main 2026MCQ

Two known resistances of $R Ω$ and $2 R Ω$ and one unknown resistance $X Ω$ are connected in a circuit as shown in the figure. If the equivalent resistance between points $A$ and $B$ in the circuit is $X Ω$ , then the value of _______ $X$

🚀 Solve in Practice Mode

Q4JEE Main 2026MCQ

Refer to the figure given below. The values of $I_1$ , $I_2$ and $I_3$ are ________.[JEE Main 5 Apr 2026 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Loop-1 $-4 ( {\text{i}}_{1}- {\text{i}}_{3}) -2 ( {\text{i}}_{1}+ {\text{i}}_{2}- {\text{i}}_{3}) - {\text{i}}_{1}+10=0$ $-7 {\text{i}}_{1}-2 {\text{i}}_{2}+6 {\text{i}}_{3}+10=0$ .....(1)Loop-2 $-5-2 ( {\text{i}}_{3}- {\text{i}}_{2}) +2 ( {\text{i}}_{1}+ {\text{i}}_{2}- {\text{i}}_{3}) =0$ $2 {\text{i}}_{1}+4 {\text{i}}_{2}-4 {\text{i}}_{3}-5=0$ ........(2)Loop-3 $- {\text{i}}_{1}+10-4 {\text{i}}_{3}-2 ( {\text{i}}_{3}- {\text{i}}_{2}) =0$ $2 {\text{i}}_{2}-6 {\text{i}}_{3}- {\text{i}}_{1}+10=0$ .....(3)On solving equation- (i) & (ii) & (iii) ${\text{i}}_{1}=2.5 {\text{A}}$ ${\text{i}}_{2}=1.875 {\text{A}}$ ${\text{i}}_{3}=1.875 {\text{A}}$

Explanation Figure 2

Q5JEE Main 2026MCQ

A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( $R_1=R_2=R_3=R_4$ ). When $R_3$ resistance is heated to some temperature, its resistance value has gone up by $10 \%$ . The potential difference $(V_a-V_b)$ (after $R_3$ is heated) is $\_\_\_\_$ v.

🚀 Solve in Practice Mode

Q6JEE Main 2026MCQ

Refer to the figure given below, current between terminals A and B is _______A.[JEE Main 5 Apr 2026 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$I=\;\frac{15}{12}=\;\frac{5}{4}=1.25$

Explanation Figure 2

Q7JEE Main 2026NAT

The heat generated in 1 minute between points $A$ and $B$ in the given circuit, when a battery of 9 V with internal resistance of $1 Ω$ is connected across these points is $\_\_\_\_$ J.

🚀 Solve in Practice Mode

Q8JEE Main 2026NAT

Two cells of emfs 1V and 2V and internal resistance 2Ω and 1Ω, respectively connected in parallel, gave current of 1A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\;\frac{\alpha }{5}A$ .The value of α is _____.[JEE Main 6 Apr 2026 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{E}}_{ {\text{eq}}}=\;\frac{\;\frac{1}{2}+\frac{2}{1}}{\frac{1}{2}+\;\frac{1}{1}}=\;\frac{\;\frac{5}{2}}{\frac{3}{2}}=\;\frac{5}{3}$ ${\text{r}}_{ {\text{Q}}}=\;\frac{3}{2}$ $i=\;{E_{ {\text{eq}}}}/{r_{ {\text{eq}}}+R}=\;\frac{\;\frac{5}{3}}{\frac{3}{2}+R}=1$ $\;\frac{5}{3}=\;\frac{3}{2}+ {\text{R}}$ ${\text{R}}=\;\frac{1}{6} Ω$ ${\text{r}}_{ {\text{eq}}}=\;\frac{3}{2}$ ${\text{E}}_{ {\text{eq}}}=\;\frac{\;\frac{2}{1}-\frac{1}{2}}{\frac{1}{1}+\;\frac{1}{2}}=1$ $i=\;\frac{E_{e q}}{r_{e q}+R}=\;\frac{1}{\frac{3}{2}+\;\frac{1}{6}}=\;\frac{3}{5} \Rightarrow \alpha =3$

Explanation Figure 1

Explanation Figure 2

Q9JEE Main 2026MCQ

The reading of the ammeter ( $A$ ) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is $\_\_\_\_$ A.

🚀 Solve in Practice Mode

Q10JEE Main 2026MCQ

The total length of potentiometer wire AB is 50 cm in the arrangement as shown in the figure. If $P$ is the point where the galvanometer shows zero reading then the length $A P$ is $\_\_\_\_$ cm.

🚀 Solve in Practice Mode

Q11JEE Main 2026MCQ

Two resistors of $100 Ω$ each are connected in series with a 9 V battery. A voltmeter of $400 Ω$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is $\_\_\_\_$ V.

🚀 Solve in Practice Mode

Q12JEE Main 2026NAT

The equivalent resistance between the points $A$ and $B$ in the following circuit is $\;\frac{x}{5} Ω$ . The value of $x$ is $\_\_\_\_$ 。

🚀 Solve in Practice Mode

Q13JEE Main 2026MCQ

A battery with EMF $E$ and internal resistance $r$ is connected across a resistance $R$ . The power consumption in $R$ will be maximum when:

🚀 Solve in Practice Mode

Q14JEE Main 2026MCQ

A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at 40 cm from the point $P$ . The null point changed to 50 cm from the point $P$ , when $16 Ω$ resistance is connected in parallel to $R_2$ . The values of resistances $R_1$ and $R_2$ are $\_\_\_\_$ .

🚀 Solve in Practice Mode

Q15JEE Main 2026MCQ

The voltage and the current between A and B points shown in the circuit are ________.[JEE Main 4 Apr 2026 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

This circuit can beHere : ${\text{E}}_{ {\text{eq}}}=\;\frac{\;\frac{27}{3}+\frac{27}{3}+\frac{27}{3}+\frac{27}{3}}{\frac{1}{3}+\;\frac{1}{3}+\;\frac{1}{3}+\;\frac{1}{3}}$ ${\text{E}}_{ {\text{eq}}}=27 {\text{V}}$ ${\text{r}}_{ {\text{eq}}}=\;\frac{3}{4} Ω$ ${\text{i}}_{ {\text{AB}}}=\;\frac{27}{6+\;\frac{3}{4}}=4 {\text{A}}$ ${\text{V}}_{ {\text{AB}}}=4 \times 6=24 Volt$

Explanation Figure 2

Q16JEE Main 2026NAT

A cylindrical conductor of length 2 m and area of cross-section $0.2 mm^2$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} m^2 / V$ . s. The value of $\alpha$ is : (electron concentration $=5 \times 10^{28} / m ^3$ and electron charge $=1.6 \times 10^{-19} C$ )

🚀 Solve in Practice Mode

Q17JEE Main 2026MCQ

A regular hexagon is formed by six wires each of resistance $r Ω$ and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

🚀 Solve in Practice Mode

Q18JEE Main 2026MCQ

In the potentiometer, when the cell in the secondary circuit is shunted with $4 Ω$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with $12 Ω$ resistance, the balance is shifted to a length of 180 cm . The internal resistance of cell is $\_\_\_\_$ $Ω$

🚀 Solve in Practice Mode

Q19JEE Main 2026MCQ

Which one of the following is not a measurable quantity?

🚀 Solve in Practice Mode

📖 Explanation

Voltage itself is not an absolute measurable quantity; only potential difference between two points can be measured. A voltmeter always measures the difference in potential between two points, not the absolute potential at a single point. Hence, "voltage difference" is measurable, but "voltage" (absolute potential) is not directly measurable. Therefore, the correct answer is voltage.

Q20JEE Main 2026NAT

When an external resistance of 5Ω is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5Ω resistor is replaced by a 2Ω resistor, a current of 0.5 A flows through it. The internal resistance of the cell is ____Ω.[JEE Main 5 April 2026 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$i=\;\frac{\epsilon }{R+r}$ $0.25=\;{\epsilon }/{5+ {\text{r}}}$ $0.5=\;{\epsilon }/{2+ {\text{r}}}$ on solving, ${\text{r}}=1 Ω$

Explanation Figure 1

Page 1 of 20Page 2 →

Want unlimited AI-generated Current Electricity questions?

Sign up free and practice with adaptive difficulty — Easy, Medium, Hard. New questions every session.

Start practising for free →