The resistance of a galvanometer coil is 25.0 Ω, and the current required for full-scale defiect

The resistance of a galvanometer coil is 25.0 Ω, and the...

Key Concepts

Ammeter Conversion Using a Shunt Resistor

To convert a galvanometer into an ammeter that can measure larger currents, a low?resistance shunt resistor is connected in parallel with the galvanometer. This allows most of the current to bypass the galvanometer so that only a small, fixed fraction passes through it, ensuring that the instrument operates at full-scale deflection when the total current is at its new maximum value.

Voltmeter Conversion Using a Series Resistor

When converting a galvanometer into a voltmeter for measuring larger voltage ranges, a resistor is added in series with the galvanometer. This series resistor drops the excess voltage such that only the fraction corresponding to the galvanometer’s full-scale deflection appears across it, thereby allowing the instrument to measure higher voltages without being damaged.

Galvanometer

A galvanometer is a sensitive current measuring device that produces a visible deflection when a current passes through its coil. It is characterized by its internal resistance and the current required to produce full-scale deflection, making it suitable for precise measurements of very small currents.

Transcript

00:01 In the given problem, we are having the resistance of a galvanometer coil and that resistance of the galvanometer coil is given as g whose value is equal to 25 .0 oom.

00:21 The current required to give a full scale deflection to the needle of this galvanometer is ig whose value has been given as 500 micro ampere or we can say this is 500 into 10 digital power minus 3 millie ampere which can also be written as 0 .5 millie ampere now in the first part of the problem we have to show in a diagram that how can we convert this galvanometer into an amometer to read a current of i is equal to 20 .0 millie ampere so in order to read this much current suppose this is the galvanometer as its needle will give a full -scale deflection a current a current of of 0 .5 milamper so definitely this much high current 20 milamper cannot be passed through this coil so what will we do we will bypass this current through a very small resistance which is known as a shunt resistance so if the current passing through this galvanometer coil is ig the remaining current means i minus ig will be passing through this shunt resistance as these two resistors means the galvanometer coil and this shunt resistance are in parallel so the potential drop taking place across these two resistances will remain same so we can say the potential drop taking place across the galvanometer coil is ig multiplied by g should be equal to the potential drop taking place across this shunt resistance so the value of this shunt resistance required will come out to be ig into g divided by i minus ig so when we we put all these known values this is ig which is 0 .5 milamper for the resistance of galvanometer coil this is 25 om then the total current which should pass through this galvanometer means which has been converted into an emometer this will be known as an m meter so the net current passing through this ammeter is i which has been given as 20 .0 millie ampere minus again this is 0 .5 millie amper so it will become 0 .5 milamper multiplied by 25 ome divided by 20 minus 0 .5 means 19 .5 millie ampere this millionpere will be cancered so the value of this shunt resistance required to be put in parallel with the galvanometer coil in order to convert it into an ammeter is given as it comes out to be 0 .64 and it becomes the answer for the first part of the problem.

04:15 Now in the second part of the problem we have to show how to convert the galvanometer to a volt meter now to read 50 500 millivolt full scale.

04:31 And we have to convert the resistance also which we require to convert it into a voltmeter so in order to convert this galvanometer coil into a volt meter a high resistance r is put in series with the galvanometer coil so that it may be able to read the required potential and that potential which we need to measure has been given as 500 this is v which is given as equal to 500 millivolt or we can say this is 0 .5 volt so the net resistance of this combination will come out to be g plus r the current required for field scale deflection is ig so using holmes law it can be given as v is to ig into net resistance which is nothing but g plus r so here this g plus r comes out to be v by ig so the value of this resistance are required to convert the galvanometer into voltmeter comes out to be v by ig minus g so now we plug in the values the known values that this is 0 .5 volt for current this is ig which means means ig is 0 .5 milanpeer...

Please give Ace some feedback