
100x−10+15x−y+10x−0=0
53x−20y=30……(1)
60y−10+15y−x+5y−0=0
17y−4x=10……(2)
on solving (1)&(2)
x=0.865
y=0.792
ΔV=0.073R=15Ω
⇒i=4.87mA
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10V is maintained across AC.

Held on 17 Mar 2021 · Verified 6 Jul 2026.
2.44μA
2.44mA
4.87mA
4.87μA
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A solenoid has a core made of material with relative permeability $400$. The magnetic field produced in the interior of solenoid is $1.0$ T. The magnetic intensity in SI units is $\alpha \times 10^5$. The value of $\alpha$ is ______. (Free space permeability $\mu_0=4\pi \times 10^{-7}$ SI units.)
Two identical small bar magnets each of dipole moment $3\sqrt{5}$ J/T are placed at a center to center separation of $10$ cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is $\alpha \times 10^{-3}$ T. The value of $\alpha$ is ______. ($\mu_0=4\pi \times 10^{-7}$ Tm/A) 
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 $1 \Omega$ is connected across these points is $\_\_\_\_$ J. 
Two known resistances of $R \Omega$ and $2 R \Omega$ and one unknown resistance $X \Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points $A$ and $B$ in the circuit is $X \Omega$, then the value of $X$ is $\_\_\_\_$ $\Omega$. 
A cylindrical conductor of length 2 m and area of cross-section $0.2 \mathrm{~mm}^{2}$ 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 $\alpha \times 10^{-3} \mathrm{~m}^{2} / V. s$. The value of $\alpha$ is : (electron concentration $=5 \times 10^{28} / \mathrm{m}^{3}$ and electron charge $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$
Work through every JEE Main Electromagnetism PYQ, year by year.