The coil in a moving coil galvanometer experi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The torque on the coil is given by $	au = NIAB \sin 	heta$. For small deflections,$	au = NIAB = C 	heta$,where $C$ is the torsional constant.G

Vedclass · Accepted Answer

$Q\sqrt {\frac{{\pi NAB}}{{2iI}}} $

Vedclass · Answer

(A) The torque on the coil is given by $	au = NIAB \sin 	heta$. For small deflections,$	au = NIAB = C 	heta$,where $C$ is the torsional constant.Given that for current $i$,the deflection is $	heta = 90^o = \frac{\pi}{2}$ radians.So,$NiAB = C \frac{\pi}{2} \Rightarrow C = \frac{2NiAB}{\pi}$ ........$(1)$When a charge $Q$ is passed suddenly,the impulse of the torque is equal to the change in angular momentum:$\int 	au dt = \int (NIAB) dt = NAB \int i dt = NABQ$Since $\int 	au dt = \Delta L = I \omega$,we have $I \omega = NABQ$ ........$(2)$By the law of conservation of energy,the initial kinetic energy is converted into potential energy of the spring at maximum deflection $	heta_{max}$:$\frac{1}{2} I \omega^2 = \frac{1}{2} C 	heta_{max}^2$$	heta_{max} = \omega \sqrt{\frac{I}{C}} = \frac{I \omega}{\sqrt{IC}} = \frac{NABQ}{\sqrt{IC}}$Substituting $C$ from $(1)$:$	heta_{max} = \frac{NABQ}{\sqrt{I \cdot \frac{2NiAB}{\pi}}} = NABQ \sqrt{\frac{\pi}{2NiABI}} = Q \sqrt{\frac{\pi NAB}{2iI}}$

Similar Questions

$A$ galvanometer has a deflection of $25$ divisions for a current of $0.1 \ A$ flowing through it. Then the current sensitivity of the galvanometer is (in $div/A$)

$A$ galvanometer has a coil of resistance $100 \, \Omega$ and gives a full-scale deflection for $30 \, mA$ current. If it is to work as a voltmeter of $30 \, V$ range,the resistance required to be added will be..... $\Omega$

$A$ galvanometer of resistance $100 \Omega$ requires $10 \mu A$ current for full-scale deflection. If a shunt resistance of $1 \Omega$ is connected in parallel to convert it into an ammeter, what is the minimum current required to obtain full-scale deflection (in $\text{ mA}$)?

The maximum current that can be measured by a galvanometer of resistance $40 \,\Omega$ is $10 \,mA$. It is converted into a voltmeter that can read up to $50 \,V$. The resistance to be connected in series with the galvanometer is ... (in $\Omega$)

Two galvanometers $A$ and $B$ require $3\,mA$ and $5\,mA$ respectively to produce the same deflection of $10$ divisions. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module