# a mili ammeter whose resistance is 120 ohm has full scale deflection in the current of 5 mili A. how will you use it to measure Maximum current find the resistance of the mili ammeter ?value of Maximum current is 50 mili A

Solution,
Given,
Resistance of the ammeter, G = 120 ohm
Maximum current for full scale deflection, I = 5 mA
In order to show full deflection for 50 mA current, A shunt resistance must be connected in parallel so that 50 - 5 = 45 mA current passes through the shunt S.
Now since the ammeter and the shunt are connected in parallel, so the voltage developed across the ammeter and the shunt S will be same.
Thus,
$5mA×120\Omega =45mA×S\phantom{\rule{0ex}{0ex}}S=\frac{5mA×120\Omega }{45mA}=\frac{120}{9}=\frac{40}{3}=13.33\Omega$

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Given, the resistance of ammeter (G) = 120 ohm
This ammeter gives full scale deflection of I= 5 mA.
Definitely, we need to shunt the ammeter by a certain value of resistance (let it be S) so as to measure the maximum current 50 mA.
Before that, let us find out the voltage across the ammeter when it has full scale deflection,
V0 = G x I
=> V0 = 120 x 5 x 10-3 V

Now, let us shunt the ammeter by a resistance of S.... (this means we will connect the resistance S in parallel to the ammeter)
So, the equivalent resistance will be = (GS/G+S) ohm
Also, the current now will be = 50 mA = 50 x 10-3 A
Hence, current through shunt = (50 x 10-3 x (GS/G+S) x (1/S)) A

Now we know, the voltage remains the same for parallel connections => hence both ammeter and shunt resistor will have same voltage.
Hence,
V0 = S x 50 x 10-3 x (GS/G+S) x (1/S)
=> 120 x 5 x 10-3 = 50 x 10-3 x (GS/G+S)
=>  12 = GS/G+S
=> 12(G+S) = GS
Value of G = 120 ohm
=> 12 (120 + S) = 120S
=> 120 + S = 10 S
=> 9S = 120
=> S = 120/9 = 40/3 = 13.33 ohm

Hence, we need to connect a shunt resistor of 13.33 ohm in parallel to the ammeter to measure a maximum current of 50mA.

Hope it helps!

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