You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Series and Parallel Combination of Resistors

A moving coil...

Question

A moving coil galvanometer, having a resistance $G$ , produces full scale deflection, when a current $I_{g}$ flows through it. This galvanometer can be converted into $(i)$ an ammeter of range $0 to I_{0} (I_{0} > I_{g})$ by connecting a shunt resistance $R_{A}$ across it, and, $(i i)$ into a voltmeter of range $0 to V (V = G I_{0})$ by connecting a series resistance $R_{V}$ with it. Then,

R_{A} R_{V} = G^{2} (\frac{I_{0} - I_{g}}{I_{g}}) and \frac{R_{A}}{R_{V}} = {(\frac{I_{g}}{I_{0} - I_{g}})}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{A} R_{V} = G^{2} and \frac{R_{A}}{R_{V}} = \frac{I_{g}}{(I_{0} - I_{g})}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{A} R_{V} = G^{2} (\frac{I_{g}}{I_{0} - I_{g}}) and \frac{R_{A}}{R_{V}} = {(\frac{I_{g}}{I_{0} - I_{g}})}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{A} R_{V} = G^{2} and \frac{R_{A}}{R_{V}} = {(\frac{I_{g}}{I_{0} - I_{g}})}^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $R_{A} R_{V} = G^{2} and \frac{R_{A}}{R_{V}} = {(\frac{I_{g}}{I_{0} - I_{g}})}^{2}$
For the ammeter,

$R_{A} = \frac{I_{g} G}{I_{0} - I_{g}}$

And, for the voltmeter,

$R_{V} = \frac{V}{I_{g}} - G = \frac{V - I_{g} G}{I_{g}}$

So, $R_{A} R_{V} = \frac{G (V - I_{g} G)}{I_{0} - I_{g}}$

$= \frac{G (G I_{0} - I_{g} G)}{I_{0} - I_{g}} = G^{2}$

And,
$\frac{R_{A}}{R_{V}} = \frac{I_{g}^{2} G}{(I_{0} - I_{g}) (V - I_{g} G)}$

$= \frac{I_{g}^{2} G}{(I_{0} - I_{g}) (G I_{0} - I_{g} G)}$

$= \frac{I_{g}^{2}}{(I_{0} - I_{g})^{2}}$

Hence, $(B)$ is the correct answer.

Suggest Corrections

Similar questions

Q. A galvanometer can be converted to a voltmeter of full-scale deflection

V_{0}

by connecting a series resistance

R_{1}

and can be converted to an ammeter of full -scale deflection

I_{0}

by connecting a shunt resistance

R_{2}

. What is the current flowing through the galvanometer at its full-scale deflection?

Q. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990

Ω

resistance, it can be converted into a voltmeter of range 0-30 V. If connected to a

\frac{2 n}{249} Ω

resistance, it becomes an ammeter of range 0 - 1.5 A. The value of

n

is :

Q. A galvanometer gives full scale deflection with

0.006 A

current. By connecting it to a

4990 Ω

resistance, it can be converted into a voltmeter of range (

0 - 30 V

). If connected to a

\frac{2 X}{249} Ω

resistance, it becomes an ammeter of range (

0 - 1.5 A

). The value of

X

is ______.

Q. A moving coil galvanometer of

50 Ω

resistance gives full deflection when a current of

0.5 mA .

To convert this galvanometer into a voltmeter of

10 V

range, the resistance to be connected in series with the galvanometer is

Q. A galvanometer has a resistance G and a current

i_{g}

flowing in it produces full scale deflection.

S_{1}

is the value of shunt, which converts it into an ammeter of range 0-I and

S_{2}

is the value of shunt for the range 0-2I the ratio

S_{1}

S_{2}

Join BYJU'S Learning Program

COURSES

EXAMS

CLASSES

EXAM PREPARATION

RESOURCES

COMPANY

FOLLOW US

FREE TEXTBOOK SOLUTIONS

STATE BOARDS

FOLLOW US

The correct option is D RARV=G2 and RARV=(IgI0−Ig)2For the ammeter, RA=IgGI0−Ig And, for the voltmeter, RV=VIg−G=V−IgGIg So, RARV=G(V−IgG)I0−Ig =G(GI0−IgG)I0−Ig=G2 And, RARV=I2gG(I0−Ig)(V−IgG) =I2gG(I0−Ig)(GI0−IgG) =I2g(I0−Ig)2 Hence, (B) is the correct answer.

COURSES

EXAMS

CLASSES

EXAM PREPARATION

RESOURCES

COMPANY

FOLLOW US

FREE TEXTBOOK SOLUTIONS

STATE BOARDS

FOLLOW US