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

Related Questions

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1.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

2.
What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.
A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

5.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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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

1. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

2.What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

5.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

2.
What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

4.
A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

5.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: