# A multi-range cur

The internal resistance of the galvanometer is 10ohms.

So the voltage drop across it is V=10-2V =10mV

Applying KVL for 10mA. [1mA through galvanometer and 9mA through s1+s2+s3]

10=9(s1+s2+s3)

(s1+s2+s3) =10/9 eq.1

In case of 100mA (1mA through galvanometer+ s1 and 9mA through s2+s3)

99(s2+s3) =10+s1

(s2+s3) =(10+s1)/99 eq.2

In case of 1A (1mA through galvanometer+ s1+ s2 and 9mA through s3)

999s3=10+s1+s2 eq.3

Putting eq 2 in Eq.1;

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