Question

A moving coil galvanometer when shunted with a resistance of 5 Ω gives a full scale deflection for 250 mA and when a resistance of 1050 Ω is connected in series, it gives a full scale deflection for 25 volt. Find the resistance of the galvanometer​

Answer 1

Answer : The answer is to put a parallel resistor to the two terminals of the galvanometer. Now lets calculate the value of the resistor needed in order to shunt the galvanometer to read 2A at full scale. We know the first thing, is that the voltage across the 2 terminals needs to be V = IR or V = 0.015 Amps * 5 ohms = 0.075 Volts.

So now we know that when the current is 2 Amps, 0.015 Amps is flowing through the galvanometer and the rest is flowing through the resistor. This indicates that the current through the resistor is 2 Amps - 0.015 Amps = 1.985 Amps.

So what is the value of the shunt resistor? It has a Voltage drop of 0.075 Volts, and a current of 1.985 Amps, so the value of the resistor is R = V/I = 0.075 Volts / 1.985 Amps = 0.03778 Ohms.

Now just to be complete, how many Watts does this resistor produce at 1.985 Amps. W = (I ^ 2) * R = (1.985 Volts ^ 2) * .03778 Ohms = 0.1488 Watts

Explanation:

Answer 2

Answer:

The resistance of the galvanometer is 50Ω

Explanation:

Given that,

A moving coil galvanometer when shunted with a resistance of 5Ω gives a full scale deflection for 250mA and when a resistance of 1050Ω is connected in series, it gives a full scale deflection for 25V.We are required to find the resistance of the galvanometer.

Let $i_{g}$ be the maximum current that can flow through galvanometer and,

R=Resistance of galvanometer,$r_{s}$=Resistance of series connected

As the full scale deflection voltage is 25V,We have-

$i_{g}(R+r_{s})=25= > i_{g}=\frac{25}{R+1050} -- > (1)$

We also have another relation for current in the galvanometer which is

$i_{g}R+i_{g}r_{p}=250\times10^{-3}r_{p}\\= > i_{g}=\frac{0.25r_{p}}{R+r_{p}} -- > (2)$

Equating both the equations 1 and 2,we get-

$\frac{25}{R+1050} =\frac{0.25\times5}{R+5} \\= > 25R+125=1.25R+1312.5\\= > 23.75R=1187.5 = > R=50\Omega$

Therefore,the resistance of the galvanometer is 50Ω

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