If the galvanometer current is 10 mA- resistance of the galvanometer is 40 - and shunt of 2 - is connected to the galvanometer- the maximum current which can be measured by this ammeter is the

The correct option is D 0.21 A i_g=10 mA, galvanometer resistance R_g=40 #937; and shunt, s= 2 #937; I= ( S+GS )I_g = (2+402 ) ...

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

Byju's Answer

Standard XII

Physics

Voltmeter

If the galvan...

Question

If the galvanometer current is 10 mA, resistance of the galvanometer is 40 $Ω$ and shunt of 2 $Ω$ is connected to the galvanometer, the maximum current which can be measured by this ammeter is the

0.21 A

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

2.1 A

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

210 A

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

21 A

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

Open in App

Solution

The correct option is D 0.21 A
$i_{g} = 10 m A, g a l v a n o m e t e r r e s i s t a n c e R_{g} = 40 Ω a n d s h u n t, s = 2 Ω$

$I = (\frac{S + G}{S}) I_{g}$ = $(\frac{2 + 40}{2})$ X $0.01 = 0.21 A$

Suggest Corrections

Similar questions

Q. In the given circuit, a galvanometer with a resistance of 70

Ω

is converted to an ammeter by a shunt resistance of 0.05

Ω

, the total current measured by this device is

Q. To convert a galvanometer into an ammeter a small resistance

S

(called the shunt) is connected in parallel with a galvanometer. A galvanometer has resistance

G

and full scale deflection current

i

. To convert this galvanometer into an ammeter of range 10 i, a shunt

S = \frac{G}{9}

is connected in parallel with G.

Now we want to measure the potential difference with the help of this ammeter. The maximum value of potential difference which can be measured with the help of this is:

Q. A moving coil galvanometer is converted into an ammeter reads up to

0.03 A

by connecting a shunt of resistance

4 r

across it and ammeter reads up to

0.06 A

, when a shunt of resistance r is used. What is the maximum current which can be sent through this galvanometer if no shunt is used ?

Q. A galvanometer shows a reading of

0.65 mA

. When a galvanometer is shunted with a

4 Ω

resistance, the deflection is reduced to

0.13 mA

. If the galvanometer is further shunted with a

2 Ω

wire, the new reading will be (the main current remains the same)

Q. The resistance of a galvanometer is

999 Ω

. It is shunted by

1 Ω

resistance. If the current in the circuit is

10^{- 2} A,

then the current through galvanometer is

Join BYJU'S Learning Program

If the galvanometer current is 10 mA, resistance of the galvanometer is 40Ω and shunt of 2Ω is connected to the galvanometer, the maximum current which can be measured by this ammeter is the

The correct option is D 0.21 Aig=10mA,galvanometer resistanceRg=40Ω andshunt,s=2ΩI=(S+GS)Ig = (2+402) X 0.01=0.21A

If the galvanometer current is 10 mA, resistance of the galvanometer is 40 $Ω$ and shunt of 2 $Ω$ is connected to the galvanometer, the maximum current which can be measured by this ammeter is the

The correct option is D 0.21 A
$i_{g} = 10 m A, g a l v a n o m e t e r r e s i s t a n c e R_{g} = 40 Ω a n d s h u n t, s = 2 Ω$

$I = (\frac{S + G}{S}) I_{g}$ = $(\frac{2 + 40}{2})$ X $0.01 = 0.21 A$