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Question: A galvanometer of resistance \(100\,\Omega \) gives a full scale deflection at \(10\,mA\) current. W...

A galvanometer of resistance 100 Ω100\,\Omega gives a full scale deflection at 10 mA10\,mA current. What should be the value of shunt so that it can measure a current of 100 mA100\,mA ?
A. 11.11  Ω11.11\,\,\Omega
B. 1.1  Ω1.1\,\,\Omega
C. 9.9 Ω9.9\,\Omega
D. 4.4 Ω4.4\,\Omega

Explanation

Solution

Understand the construction and working principle of the galvanometer. Shunt plays an important role in converting galvanometer into ammeter.Galvanometer converts electrical signals to mechanical work by moving the needle.A full scale deflection current refers to the maximum current that is measured by the galvanometer.

Complete step by step answer:
A galvanometer is an electromechanical device used to measure the magnitude and direction of current. It is a very sensitive device which can measure currents of the order of a few micro amperes. It mainly consists of a rectangular coil made up of thinly insulated copper wire. This wire is wounded on a metallic frame and the coil is free to move about a fixed axis.

In order to make the field radial, a soft cylindrical iron core is inserted in the coil. The principle followed by a galvanometer is that a magnetic torque is experienced by a current-carrying conductor when placed in an external magnetic field. This magnetic torque deflects the coil such that it twists at an angle. This angle through which the coil deflects is directly proportional to the magnitude of the current in the coil.

As mentioned, a galvanometer is a sensitive device which can measure only small currents.To convert a galvanometer into an ammeter so that it can measure a large current, a low resistance shunt is connected parallel to the galvanometer. The maximum current passes through the shunt. The value of shunt is determined by the amount of current which is to be measured.

Let, Rg={{R}_{g}}= galvanometer resistance, Ig={{I}_{g}}= current passing through the galvanometer producing full deflection, Rs={{R}_{s}}= shunt resistance, Is={{I}_{s}}= shunt current, I=I= total current in the circuit and V=V= voltage.The given quantities are provided in the question:
Rg=100 Ω{{R}_{g}}=100\,\Omega , Ig=10 mA{{I}_{g}}=10\,mA , I=100  mAI=100\,\,mA

As the shunt and galvanometer are parallel , voltage must be equal. The voltage in the galvanometer is given by,
V=IgRg ⇒V=10×10−3A×100 Ω V={{I}_{g}}{{R}_{g}} \\\ \Rightarrow V=10\times {{10}^{-3}}A\times 100\,\Omega \\\
I=Is+Ig ⇒Is=I−IgI={{I}_{s}}+{{I}_{g}} \\\ \Rightarrow {{I}_{s}}=I-{{I}_{g}}

The voltage equation in the shunt is given by:
V=IsRs ⇒V=(I−Ig)Rs ⇒V=(100×10−3−10×10−3)A×Rs V={{I}_{s}}{{R}_{s}} \\\ \Rightarrow V=\left( I-{{I}_{g}} \right){{R}_{s}} \\\ \Rightarrow V=\left( 100\times {{10}^{-3}}-10\times {{10}^{-3}} \right)A\times {{R}_{s}} \\\
Equating both the voltages:
(100×10−3−10×10−3)×Rs=10×10−3×100 ⇒90×10−3×Rs=10×10−3×100 ⇒Rs=10×10−3×10090×10−3 ⇒Rs=11.11Ω\left( 100\times {{10}^{-3}}-10\times {{10}^{-3}} \right)\times {{R}_{s}}=10\times {{10}^{-3}}\times 100 \\\ \Rightarrow 90\times {{10}^{-3}}\times {{R}_{s}}=10\times {{10}^{-3}}\times 100 \\\ \Rightarrow {{R}_{s}}=\dfrac{10\times {{10}^{-3}}\times 100}{90\times {{10}^{-3}}} \\\ \Rightarrow {{R}_{s}}=11.11\Omega
The shunt resistance is 11.11 Ω11.11\,\Omega .

Hence option A is correct.

Note: Though galvanometer and ammeter are both used to measure current but there is a difference in their nature of measurement.A Galvanometer can measure only direct current but also specifies the direction of the current. On the other hand, an ammeter can measure both alternating current as well as direct current.