A galvanometer having resistance 100 ohms shows full scale deflection with current 10 mA. Find value of shunt to convert it into an ammeter of 10 ampere range.

A galvanometer can be defined as a device used for the detection of small electric current or to measure its magnitude. In general, the current and its intensity is shown by the movement of a magnetic needle or a coil in a magnetic field. When current I_Gflows via the galvanometer, the current through the shunt resistance can be expressed by I_S= I – I_G.The voltages across the galvanometer and shunt resistance are equivalent to one another due to the parallel nature of their connection.

Therefore, R_G.I_G= (I – I_G).R_s

Here,

R_G= Resistance of galvanometer
G = Galvanometer coil
I = Total current via circuit
I_G = Total current via galvanometer that corresponds to full-scale reading
R_s = Value of shunt resistance

Step-1: Solution

As per the given question, the following can be determined:

I = 10A
Ig = 10 mA = 0.010 A
Rg = 100 Ω

Thus, with the given condition, we can obtain:

I = I_g+ I_s

⇒ (I − I_g)R_s= I_gR_g

By replacing the known values, we can get:

(10 − 0.010)Rs = 0.010 × 100

⇒ R_s= 0.1

Also Read: NCERT Solutions for Class 6 to 12

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A galvanometer having resistance 100 ohms shows full scale deflection with current 10 mA. Find value of shunt to convert it into an ammeter of 10 ampere range.

Step-1: Solution

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3.
Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

4.
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Comments

dOGVmW

A galvanometer having resistance 100 ohms shows full scale deflection with current 10 mA. Find value of shunt to convert it into an ammeter of 10 ampere range.

Step-1: Solution

Related Questions

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2.Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

3.Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

4.The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

5.The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

6.A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is \( \sqrt{2} \), and \( \sqrt{3} \).

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Comments

dOGVmW

2.
Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

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Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

4.
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