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Question

A galvanometer is used to measure small currents. A certain galvanometer has a resistance of 500Ω and gives a full-scale deflection for a current of 200μA. This meter is connected as shown in the figure to make a multirange current meter.
Connection to the circuit is made at the terminals shown. The currents in the external circuit needed to give full-scale deflections when X is connected to A,B and C, in turn, are shown in the table.

X connected toCurrent in the external circuit (mA)
A1
B10
C100
Find out the of R3 is
744093_d358d0879e254cd1960e615ae388a38a.png

A
2.25Ω
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B
0.25Ω
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C
1.25Ω
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D
3.25Ω
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Solution

The correct option is C 1.25Ω
Given: A certain galvanometer has a resistance of 500Ω and gives a full-scale deflection for a current of 200μA. This meter is connected as shown in the figure to make a multi-range current meter.
Connection to the circuit is made at the terminals shown. The currents in the external circuit needed to give full-scale deflections when X is connected to A,B and C, in turn, are shown in the table.
To find R3
Solution:
The given circuit can be re-written as shown in fig(i),
Now applying Kirchhoff's Law, we get
500×200μA=(R1+R2+R3)i1500×200×106A=(R1+R2+R3)[1ma200μA]R1+R2+R3=10×104×106106[1000200]R1+R2+R3=125....(i)
Now, according to the given criteria, when X is connected to B the circuit becomes as shown in fig(ii), and we get
(500+R1)200μA=(R2+R3)[10mA200μA](500+R1)200×106=(R2+R3)106[10000200]500+R1=49(R2+R3)....(ii)
Again when X is connected to C the circuit becomes as shown in fig(iii), and we get
(500+R1+R2)200μA=(R3)[100mA200μA](500+R1+R2)200×106=(R3)106[100000200]500+R1+R2=499(R3)....(iii)
From (i) and (iii),
500+125R3=499R3500R3=625R3=1.25Ω
is the required value.

931939_744093_ans_2d4c936d54924e7aa61f7f2367595785.png

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