Conversion of Galvanometer into an Ammeter

What is Ammeter?

An Ammeter is an instrument that is used to measure the electric current in the electric circuits directly in Ampere. The instrument which measures the current of the order of milliampere $(mA)$ is called the milliammeter. The internal resistance of the ideal ammeter is always zero.

What is Galvanometer?

The galvanometer is an instrument that is used to measure the very small amount of the electric charge passing through the circuit. The internal resistance of the Galvanometer is not zero.

Conversion of Galvanometer into an Ammeter

When a Galvanometer is used in the circuit and connected in the series to measure the electric current: The galvanometer is used in series to measure the electric current of the circuit so that the whole amount of the current passing through it. but the galvanometer will have some resistance due to the resultant resistance of the circuit increasing and the current in the circuit somewhat decreasing. Therefore the current read by the Galvanometer is less than the actual current.

Galvanometer used as Ammeter: To use the galvanometer as a metre in the circuit, The resistance of the galvanometer should be very small or almost zero as compared to the other resistance of the circuit. Because the internal resistance of an ideal ammeter is zero.

So a low resistance is connected in parallel to the galvanometer which is known as a shunt.

When a low resistance is connected in parallel to the galvanometer then the resultant resistance decreases as compared to the other resistance of the circuit and it can be easily used as an Ammeter and the actual current can measure through it.

Let us consider, G is the resistance of the coil of the Galvanometer and the $i_{g}$ current passing through it. If $i$ is the maximum current of the circuit then a part of current $i_{g}$ passes through the galvanometer and the remaining current $(i-i_{g})$ passes through the shunt $S$. Since $S$ and $G$ are parallel, the potential difference across them will be the same:

$i_{g} \times G = \left( i- i_{g} \right) \times S \qquad(1)$

$\frac{i_{g}}{i}=\frac{S}{S+G}$

i.e only $\frac{S}{S+G}$th part of the total current will flow in the coil of the Galvanometer. Again from equation $(1)$:

$S=\left(\frac{i_{g}}{i-i_{g}}\right)G \qquad(2)$

If the current $i_{g}$ passes through the coil of galvanometer and produces a full-scale deflection on meter scale of galvanometer, then the current $i$ in the circuit corresponds to the full-scale deflection. Thus, with a shunt $S$ of the above value, the galvanometer will be an ammeter in the range $0$ to $i$ ampere. Example: Let a current of $1 A$ in the coil of a galvanometer produces a full-scale deflection. To convert it into an ammeter of range $10A$, a shunt is required such that when the current in the circuit is $10A$, only $1A$ flows in the coil remaining passes through the shunt. Now From substitute the $i_{g}=1A$ and $i=10A$ in the above equation $(2)$:

$\frac{S}{G}= \frac{1}{\left( 10 -1 \right)}$

$\frac{S}{G}= \frac{1}{9}$

The resistance of the shunt should be only $\frac{1}{9}$th the resistance of the galvanometer coil.

Note: As the shunt resistance value is very small so the combined resistance of the galvanometer and the shunt also becomes very small and hence the ammeter has a much smaller resistance than the galvanometer.

Resistance of Ammeter:

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

$R=\frac{GS}{G+S}$

Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

click here »

Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

click here »

Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ $\frac{d^{2} \psi(x)}{d x

click here »

Physics Vidyapith ✍️

Search This Blog