In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be [2014]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be [2014]

1 $\frac{1}{499} G$

2 $\frac{499}{500} G$

3 $\frac{1}{500} G$

4 $\frac{500}{499} G$

Ans.
(3)

Here, resistance of the galvanometer $= G$

Current through the galvanometer,

$I_{G} = 0.2 % of I = \frac{0.2}{100} I = \frac{1}{500} I$

$∴$ Current through the shunt,

   $I_{S} = I - I_{G} = I - \frac{1}{500} I = \frac{499}{500} I$

As shunt and galvanometer are in parallel

$∴$ $I_{G} G = I_{S} S$

   $(\frac{1}{500} I) G = (\frac{499}{500}) S or S = \frac{G}{499}$

Resistance of the ammeter $R_{A}$ is

   $\frac{1}{R_{A}} = \frac{1}{G} + \frac{1}{S} = \frac{1}{G} + \frac{1}{\frac{G}{499}} = \frac{500}{G}; R_{A} = \frac{1}{500} G$

Want Us to Email you About Special Offers & Updates?