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

Q321JEE Main 2017MCQ

Which of the following statements is false ?[Main 2017]

🚀 Solve in Practice Mode

📖 Explanation

(1) On interchanging Cell & Galvanometer.(2) on balancing condition $R_1/R_3=R_2 /R_4$ ......(1) on balancing condition $R_1/R_2=R_3/R$ As we see both equation (1) & (2) are same. So 4th statement is false.

Explanation Figure 1

Explanation Figure 2

Explanation Figure 3

Q322

JEE Main 2017

MCQ

A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key $K_3$ is open. The balance point is at a length $l_1$ cm from P when two way key $K_3$ is plugged in between 2 and 1, while the balance point is at a length $l_2$ cm from P when key $K_3$ is plugged in between 3 and 1. The ratio of two resistances $R_1/R_2$ , is found to be :[Main 8 April 2017]

🚀 Solve in Practice Mode

📖 Explanation

When key is at point $V_1=iR_1=xl_1$ when key is at (3) $V_2=i(R_1+R_2)=xl_2$ $R_1/{R_1+R_2}=l_1/l_2$ $R_1/R_2=l_1/{l_2-l_1}$

Q323JEE Main 2017MCQ

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into to voltmeter of range $0-10$ V is :[Main 2017]

🚀 Solve in Practice Mode

📖 Explanation

$10=(5\times 10^{-3})(15+R)$ r = 1985 Ω

Q324JEE Main 2017MCQ

A 9 V battery with internal resistance of 0.5 Ω is connected across an infinite network as shown in the figure. All ammeters $A_1, A_2, A_3$ and voltmeter V are ideal.Choose correct statement.[Main 8 April 2017]

🚀 Solve in Practice Mode

📖 Explanation

$x=\frac{4x}{4+x}+2$ $x=\frac{8+6x}{4+x}$ $4x+x^2=8+6x$ $x^{2-}2x-8=0$ $x={2±\sqrt{4-4(1)(-8)}}/2={2±\sqrt36}/2$ $={2±6}/2=4$ $A'=9/{4+0.5}=2$ $A'=9$

Explanation Figure 2

Q325JEE Main 2017MCQ

In the above circuit the current in each resistance is :[Main 2017]

🚀 Solve in Practice Mode

📖 Explanation

Taking voltage of point A as = 0Then voltage at other points can be written as shown in figureHence voltage across all resistance is zeroHence current = 0

Explanation Figure 2

Q326JEE Main 2017MCQ

In a meter bridge experiment resistances are connected as shown in the figure.Initially resistance $P=4Ω$ and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is :[Main 9 April 2017]

🚀 Solve in Practice Mode

Q327JEE Main 2017MCQ

A uniform wire of length l and radius r has a resistance of 100 Ω. It is recast into a wire of radius $r /2$ .The resistance of new wire will be :[Main 9 April 2017]

🚀 Solve in Practice Mode

Q328JEE Main 2017MCQ

The figure shows three circuits I, II and III which are connected to a 3V battery. If the powers dissipated by the configurations I, II and III are $P_1,P_2$ and $P_3$ respectively, then :[Main 9 April 2017]

🚀 Solve in Practice Mode

Q329JEE Main 2016MCQ

A 50 Ω resistance is connected to a battery of 5 V. A galvanometer of resistance100 Ω is to be used as an ammeter to measure current through the resistance, for this a resistance $r_s$ is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of the current without the ammeter in the circuit ?[Main 9 Apr 2016]

🚀 Solve in Practice Mode

📖 Explanation

Its clear, shunt should be applied in parallel and least the shunt resistance, better the ammeter is.

Q330JEE Main 2016MCQ

In the circuit shown, the resistance r is a variable resistance. If for $r=fR$ , the heat generation in r is maximum then the value of f is :[Main 9 Apr 2016]

🚀 Solve in Practice Mode

📖 Explanation

to get maximum heat generation from r. $r=R_{eq}=R/2$

Explanation Figure 2

Q331JEE Main 2016MCQ

A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving afull scale deflection for a current of 10 A is :[Main 2016]

🚀 Solve in Practice Mode

📖 Explanation

For full scale deflection $100\times i_g=(i-i_g)$ Swhere 'S' is the required resistance $S={100\times 1\times 10^{-3}}/(10-10^{-3})$ $S\approx 0.01\Omega$

Q332JEE Main 2016MCQ

A galvanometer has a $50$ division scale.Battery has no internal resistance. It isfound that there is deflection of $40$ divisions when $R=2400Ω$ . Deflection becomes 20 divisions when resistance taken fromresistance box is $4900Ω$ . Then we can conclude :[Main 10 Apr 2016]

🚀 Solve in Practice Mode

Q333JEE Main 2016MCQ

The resistance of an electrical toaster has a temperature dependence given by $R(T)=R_0[1+\alpha (T-T_0)]$ in its range of operation.At $T_0=300K,R=100\Omega$ and at $T=500K,R=120\Omega$ .The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is :[Main 10 Apr 2016]

🚀 Solve in Practice Mode

📖 Explanation

$(200)^2/{R_0(1+\alpha (T-T_0))}$ $T\to$ temperature at 't' $T_0\to$ temperature at $t=300K$ $T-T_0={500-300}/30(t)$ $T-T_0=200/30t$ $T-T_0={20t}/3$ $∫^{30}_{0}(200)^2/{100(1+\alpha {20t}/3)}dt={200\times 200}/100∫^{30}_{0}\frac{dt}{1+{20\alpha }/3t}$ $=\frac{400\times 3}{20\alpha }\text{ln}({{1+20\alpha }/3\times 30}/1)$ $120=100(1+\alpha (200))$ $1+(200)\alpha =6/5$ $(200\alpha )=1/5$ $\alpha =1/1000$ $=60,000\text{ln}(6/5)$

Q334JEE Main 2015MCQ

In the circuit shown, the current in the 1Ω resistor is : [Main 2015]

🚀 Solve in Practice Mode

📖 Explanation

Taking the potential at Q to be 0 and at P to be V, weapply Kirchoff's current law at Q: ${V+6}/3+V/1+{V-9}/5=0$ $C=-3/23=-0.13volt$ The current will flow from Q to P.

Explanation Figure 2

Q335JEE Main 2015MCQ

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is $2.5 \times 10^{-4} ms^{-1}$ .If the electron density in the wire is $8 \times 10^28 m^{-3}$ ,the resistivity of the material is close to[Main 2015]

🚀 Solve in Practice Mode

📖 Explanation

$J=nev_d$ $\frac{A\Delta V}{\rho lA}=nev_d$ $\therefore \rho =\frac{\Delta V}{lnev_d}=5/{0.1\times 8\times 10^28\times 1.6\times 10^{-19}\times 2.5\times 10^{-4}}$ $=1.56\times 10^{-5}$ $\approx 1.6\times 10^{-5}\Omega m$

Q336JEE Main 2014MCQ

In a large buildings,there are 15 bulbs of 40 W,5 bulbs of 100 W,5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220V. The minimum capacity of the main fuse of the building will be:[Main 2014]

🚀 Solve in Practice Mode

📖 Explanation

ItemNo.Power40 W bulb15600 watt100 W bulb5500 watt80 W fan5400 Watt1000 W heater11000 WattTotal Wattage=2500 Watt So current capacity i $=P/V=2500/220=125/11=11.36≅12$ Amp.

Q337JEE Main 2013MCQ

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.Statement- I: Higher the range, greater is the resistance of ammeter.Statement- II: To increase the range of ammeter, additional shunt needs to be used across it.[Main 2013]

🚀 Solve in Practice Mode

📖 Explanation

For Ammeter,S $=\frac{I_gG}{I-I_g}$ So for I to increase, S should decrease, so additional S can be connected across it.

Q338JEE Main 2013MCQ

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?[Main 2013]

🚀 Solve in Practice Mode

📖 Explanation

Resistance of bulb $= {120\times 120}/60=240\Omega$ Resistance of Heater $={120\times 120}/240=60\Omega$ Voltage across bulb before heater is switched on, $V_1=120/246\times 240$ Voltage across bulb after heater is switched on, $V_2=120/54\times 48$ Decrease in the voltage is $V_1-V_2=10.04$ (approximately)Note: Here supply voltage is taken as rated voltage.

Explanation Figure 1

Q339JEE Main 2012MCQ

Two electric bulbs marked $25 {\text{W}}-220 {\text{V}}$ and $100 {\text{W}}-220 {\text{V}}$ are connected in series to a $440 {\text{V}}$ supply. Which of the bulbs will fuse?

🚀 Solve in Practice Mode

📖 Explanation

The current upto which bulb of marked $25 W-220 {\text{V}}$ , will not fuse $I_1=\;\frac{W_1}{V_1}=\;\frac{25}{220} A m p$ Similarly, $I_2=\;\frac{W_2}{V_2}=\;\frac{100}{220} {\text{Amp}}$ The current flowing through the circuit $\;I=\;{440}/{ {\text{R}}_{e f f}}, {\text{R}}_{e f f}=R_1+R_2$ $\;R_1=\;\frac{V_1^2}{P_1}=\;\frac{(220)^2}{25}$ $\;R_2=\;\frac{V_2^2}{P}=\;\frac{(220)^2}{100}$ $\;I=\;\frac{440}{\frac{(220)^2}{25}+\;\frac{(220)^2}{100}}$ $\;=\;\frac{440}{(220)^2 [\;\frac{1}{25}+\;\frac{1}{100}] }$ $\;I=\;\frac{40}{220} A m p$ $\;\;\text{ as }\; I_1 (=\;\frac{25}{220} A) < I (=\;\frac{40}{220} A) < I_2 (=\;\frac{100}{200} A)$ Thus the bulb marked $25 W-220$ will fuse.

Explanation Figure 1

Q340JEE Main 2011MCQ

If a wire is stretched to make it $0.1 \%$ longer, its resistance will:

🚀 Solve in Practice Mode

📖 Explanation

Resistance of wire $R=\;\frac{\rho l}{A}=\;\frac{\rho l^2}{C} \;\text{ (where }\; A l=C \;\text{ ) }\;$ $\therefore$ Fractional charge in resistance $\;\frac{\Delta R}{R}=2 \;\frac{\Delta l}{l}$ $\therefore$ Resistance will increase by $0.2 \%$

← Page 16Page 17 of 20Page 18 →

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 →