The current that will given full scale deflection in the absence of the shunt is nearly equal to the current through the galvanometer when shunt is connected i.e. Ig As Ig=G+SIS =120+15.5×1=0.045 ampere
A shunt of resistance 1 ~Ω is connected across a galvanometer of 120 ~Ω resistance. A current of 5.5 ampere gives full scale deflection in the…
JEE Main 2013 — Physics Electromagnetism
A shunt of resistance 1 Ω is connected across a galvanometer of 120 Ω resistance. A current of 5.5 ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly :
Held on 9 Apr 2013 · Verified 6 Jul 2026.
Options
- A
5.5 ampere
- B
0.5 ampere
- C
0.004 ampere
- D
0.045 ampere
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Electromagnetism in other years
More JEE Main Electromagnetism questions
A magnetic field vector in an electromagnetic wave is represented by $\vec{B}=B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{j}$. Its associated electric field vector is ______.
A plane electromagnetic wave is moving in free space with velocity $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and its electric field is given as $\vec{E}=54 \sin (k z-\omega t) \hat{j} \mathrm{~V} / \mathrm{m}$, where $\hat{j}$ is the unit vector along $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:
A charged particle moves in a uniform magnetic field. If the velocity is perpendicular to the field, the path of the particle is:
A monochromatic source of light operating at $15$ kW emits $2.5\times 10^{22}$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take $h=6.6\times 10^{-34}$ J·s and $c=3\times 10^8$ m/s).
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma=10 \mathrm{mho} / \mathrm{m}$. The ratio of maximum conduction current density to maximum displacement current density is $\_\_\_\_$. $\left[\text{Take }\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]$
Other Physics topics for JEE Main
Keep practicing Electromagnetism
Work through every JEE Main Electromagnetism PYQ, year by year.