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

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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.
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5.
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 (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

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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

Related Questions

CBSE CLASS XII Related Questions

2.A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

3.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

5.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 (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

5.
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 (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.