I am 12th grader just in case. Anyway, I do understand how a galvanometer is converted into an ammeter but it's logic doesn't seem to add up. This is what I know so far. Galvanometer is too sensitive therefore is modified to be used to measure current instead of only detect it. The current for full scale deflection through it is lets say Ig. Its resistance, Rg. Now, a shunt resistance is added in parallel. The maximum current that the ammeter produced can measure is lets say I and shunt resistance is Rs. Now Rs is of such a value that Ig passes through the galvanometer and the remaining current (I-Ig) passes through the shunt resistance. I understand why the shunt resistance has to be of a smaller value and the resistance of voltmeter has to be of a really greater value but what I fail to comprehend is how can we successfully measure current when the deflection of the needle is going to always be maximum due to the passage of Ig current (which is the current that causes maximum deflection). Ig is always passing through the galvanometer so how does the ammeter really work and give us current in amperes?
2 Answers
Let's look at a very simple example first. We have a galvanometer that has a full-scale reading of \$ I_G \$ and we want to measure currents up to four times that value.
simulate this circuit – Schematic created using CircuitLab
Figure 1. By making \$ R_S = \frac {1}{3}R_G \$ three quarters of the current will be handled by the shunt and the remaining quarter by the galvanometer.
Note that whatever current is flowing in the circuit being measured that the current will always split 3:1 due to the ratios of the shunt and galvanometer resistances.
If we need to measure currents of 100 times the galvanometer's max rating then we need a shunt of \$ R_S = \frac {1}{99}R_G \$ to divert 99% of the current from the galvanometer.
I understand why the shunt resistance has to be of a smaller value and the resistance of voltmeter has to be of a really greater value but what I fail to comprehend is how can we successfully measure current when the deflection of the needle is going to always be maximum due to the passage of Ig current
Remember that the galvanometer doesn't "hog" all the current if the current is a low value. The current always splits inversely to the ratio of the resistances.
how can we successfully measure current when the deflection of the needle is going to always be maximum due to the passage of Ig current (which is the current that causes maximum deflection)
You understood this part wrong. Maximum deflection occurs at Ig(max). If Ig is less than Ig(max), the needle will stay somewhere in between.
Deflection of needle is not always going to be maximum.
When you are designing the ammeter, you need to select a value of Rs depending upon the max current you intend to measure. For ex - an ammeter which can measure current up to 1 A will use a larger value of Rs as compared to an ammeter which can measure a current up to 100 A.
For the ammeter rated for 1 A, the Ig(max) value will be reached only when you are allowing 1 A through the ammeter. For the ammeter rated 100 A, Ig(max) will be reached when measured current is 100 A.
For current values lesser than rated values, the current through galvalometer will be less than Ig(max) and hence you will be able to measure the current through the ammeter. If you pass a current higher than rated value (eg - 2 A through 1 A rated ammeter), the needle will show full deflection and there is no way of knowing the exact value of current passing through it.
