Converting a galvanometer to ammeter and voltmeter.

AI Thread Summary
To convert a galvanometer with a resistance of 12 ohms and full-scale deflection at 3 mA into an ammeter with a range of 0 to 6 A, additional resistors must be added in parallel to ensure minimal disturbance to the circuit's current. For the voltmeter conversion to a range of 0 to 18 V, a high resistance must be used to ensure that the voltmeter does not draw significant current, allowing accurate voltage measurement. The design requires careful calculation of resistor values to achieve the desired ranges while maintaining the galvanometer's integrity. Users are encouraged to attempt solving these problems independently for better understanding, as simply copying answers does not facilitate learning. Engaging with the concepts of circuit design and resistance is crucial for successful conversions.
a.gracias
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How will you convert a galvanometer of resistance 12 ohms showing full scale deflection for a current of 3 milli ampere to
(i) Ammeter of range o to 6 Ampere
(ii) Voltmeter of range 0 to 18V.
 
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a.gracias said:
How will you convert a galvanometer of resistance 12 ohms showing full scale deflection for a current of 3 milli ampere to
(i) Ammeter of range o to 6 Ampere
(ii) Voltmeter of range 0 to 18V.
Please follow the posting template -- it's there for a reason.

We don't do your homework for you here (it's against forum rules); You must show some attempt to solve the problem so that we can see how to help you to solve it yourself.
 
gneill said:
Please follow the posting template -- it's there for a reason.

We don't do your homework for you here (it's against forum rules); You must show some attempt to solve the problem so that we can see how to help you to solve it yourself.

But I don't know how to crack this question.
Any help will be appreciated
Pls help, I plead.
 
What other electrical quantity is known as far as the galvanometer goes when it is at full deflection?
 
Well, this is a very easy question, and you should really try to solve it for yourself. You don't learn much from just copying our answers here. Just some hints for the solution:

(a) Ammeter
---------------

You want to measure the current by plugging the Ammeter in the circuit of which you want to measure the current such that you disturb this current as little as possible. That means, in your extension of the available range you should plug in resistors in addition to the galvanometer such that the resistance of the so obtained ammeter is as small as possible (it's good to think about the question, why this must be so). So think, how you can achieve this and then calculate the value of the needed resistor.

(b) Voltmeter
---------------

Same arguments as for the Ammeter. Only now you must aim at an as high resistance of your voltmeter as possible (think again, why!).
 
Hint for part 1): Clearly if the meter reads full scale at 3mA you can't send 6A through it. So you need to design a circuit that fits between the two meter probes (lets call them A and B) so that when 6A is flowing between A and B only 3mA of it is going through the meter, the rest is taking another path between A and B.
 

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