# How does galvanometer measure more current than is passing through it?

My textbook clearly states:

after putting a shunt in parallel to it, a galvanometer becomes an ammeter.

The diagram is quite similar to this:

This is fine. I have problem with this:

Now the same acale of the galvanometer which was recording the maximum current $I_g$ before conversion into ammeter will record maximum current I after conversion into ammeter. It means each division of scale in ammeter will be showing higher current than that of galvanometer.

I don't understand how can galvanometer measure current which is not even passing through it? I know we can calculate I from Ig as:

$I_g = I \times S \div (G+S)$

So if we know Ig (which galvanometer is measuring) we can find I. But how does that mean that galvanometer's scale changes and it starts measuring the current which is not even paasing through it?

• What are S and G? And have you learned the "current divider rule"? – The Photon Feb 22 '17 at 5:55
• S=shunt resistance, G=galvanometer resistance. You mean I=Ig+Is (in this case)? – completely newbie Feb 22 '17 at 6:01
• No, the current divider rule. How to figure out how much current goes through each of two resistors when you put them in parallel. – The Photon Feb 22 '17 at 6:08
• Hint: If you know S and G, and you know the potential across them is the same because they're connected in parallel, then you can find the ratio between $I_g$ and $I_s$. – The Photon Feb 22 '17 at 6:09
• Is×S=Ig×G? If not, then I don't know about it. – completely newbie Feb 22 '17 at 6:11

Galvanometer still shows maximum deflection when maximum current is passing through it. But out of the total current in the circuit only some of it goes towards galvanometer and rest goes to shunt. So the actual current in the circuit is more than which is passing through galvanometer.

But the galvanometer is calibrated accordingly that it will show the reading of $I$ and not $I_g$.

Galvanometer still shows deflection proportional to $I_g$ but the reading is different as the calibration is different. Hence the reading is I though current in galvanometer though the current through it is Ig because the 1 divisions or marking on ammeter is actually different than what actual current is passing through it.

So though $I_g$ is current it shows $I_g+I_s$ as it is calibrated as such.

So suppose there are 10 divisions then Ig is maximum current so when Ig/10 current passes it shows deflection of one division but the galvanometer is calibrated as 1 division =I/10. That maximum current measured by ammeter is I.

• Is this what you wanted?? – ATHARVA Feb 22 '17 at 6:27
• Yes calibration is the point I was looking for. Can you add some more about calibration in galvanometer? – completely newbie Feb 22 '17 at 6:34
• It is similar to that how weighing machine weighs tells our mass. Though a weighing machine measures your weight it is calibrated ti give us a value whichever is 1/9.8 times our weight. So it gives rereading of our mass – ATHARVA Feb 22 '17 at 6:39
• So if you take a weighing machine on moon to weigh you will get wrong readings as it is calibrated with respect to earth's gravity. Similar thing is with galvanometer it actually measures Ig but shows you I.i will add something more in the answer – ATHARVA Feb 22 '17 at 6:41

To make it more clear, redraw your schematic with the galvonometer drawn as a resistance (with value G) instead of as a meter.

Now, you know the potential across the two branches (the shunt resistor and the galvonometer) is equal because they're connected in paralllel.

And you know the equivalent resistance of each branch (because you've named them "S" and "G").

From that you can find the ratios between $I_g$, $I_s$, and $I$. Which will lead you to the result you're trying to get.

• That is okay, but that is not what I asked. If Ig is passing through galvanometer, then how is it itself measuring Ig+Is? – completely newbie Feb 22 '17 at 6:16
• Because you know the ration $I_s/I_g$, then knowing $I_g$ tells you $I_s$. And if you then know both $I_g$ and $I_s$, then you know $I$. – The Photon Feb 22 '17 at 6:24
• So, in short, galvanometer is measuring Ig only? – completely newbie Feb 22 '17 at 6:31
• You could look at it that way, and say it's the composite instrument of the galvanometer and the shunt resistor that's measuring the whole current. But it's a matter of semantics and other people might just say "use the galvanometer to measure I". – The Photon Feb 22 '17 at 6:33

First you need to understand why we attach something called a shunt to convert a galvanometer into an ammeter after all? This is because a galvanometer is a very sensitive device. It gives a full scale deflection (maximum possible reading) for current of the order of microamperes (normally current encountered is at least of the order of milliamperes), so if you connect it directly in a circuit, the needle will reach the maximum value and start oscillating rapidly (often with a characteristic sound). Secondly a galvanometer has an appreciable resistance (~100ohms) which changes the current that was initially in circuit.

Let's assume that the current in a circuit is I just before the point of attachment of the galvanometer and the shunt. At the junction, the current will fork into two: I' and i where I' is the current through the shunt and i is the current through the galvanometer.

What you need to understand is that while doing any conversion, the value of the shunt is something we have to calculate for a particular maximum value of current we require in the circuit. What's known is only the resistance of the galvanometer and a quantity called it's figure of merit (figure of merit is just the current required for a unit deflection). What we're doing is basically calibrating the value of the current through the galvanometer such that it gives us an idea of the current through the circuit. (In your case this is what enables it to measure a current greater than what actually passes through it. It=galvanometer)

In questions you'll be given a value of current which produces a maximum deflection in the galvanometer or the figure of merit of the galvanometer and the number of divisions a readings runs into.

Getting back, it follows from Kirchhoff's junction rule that the current at the point where it 'forks off' would be I= I' + i. In other words the current in the circuit split into two at the junction. A gross oversimplification would be something like this: Charges? They're like us driving on a road. We prefer to take the least clogged lanes. Much like us, charges will take the path which is the least clogged, i.e. the path which offers the least resistance. Here that path is through the shunt resistance.

So that does it for the stuff that happens at the junction. Now since the shunt and galvanometer are connected in parallel, the voltage drop across both would be the same. Mathematically, I'S= iG where S and G are the shunt and galvanometer resistance respectively. Also, I'= I - i Which makes us arrive at the formula, S= iG÷ (I - i)

Having rambled on for a few paragraphs, I realise that that might not have answered your doubt. I'll try to rephrase. Connecting two resistances in parallel we achieve a net resistance which is smaller than the smallest value of resistance connected in the combination. That's the case with the present scenario. We want a very small resistance introduced into the circuit when we add the galvanometer and that's why we add the shunt. Next, by the traffic analogy, the current which is flowing through the galvanometer is much smaller than what flows through the shunt (current which flows through the shunt, can be said to be very close to the value of the current which flows through the circuit). The formula establishes a relationship between the current which flows through both and calibrates the galvanometer to read the current through the circuit (current through the circuit is very very very close to the current through the shunt!) And this is how we're able to measure a value of current which is much greater than what actually passes through the galvanometer

Note: Though I haven't said much about the figure of merit, it's pretty much what the entire conversion revolves around.

Hope that helps.