Answer :

0.15A, 0.83A

Given,


The resistance of the rheostat, R= 30Ω


The emf of the battery, E=5.5V


Formula used:


The rheostat can vary the resistance from 0Ω to maximum, and will be added as a series resistance to the given setup.


Solution:


The 10 Ω and 20 Ω connected parallel to each other. This can be reduced to a single resistance as they both connected to same potential difference. So the effective resistance, Reff between 10 Ω and 20 Ω will be,



The rheostat resistance will be added in series to the above resistance.


The minimum current will be marked when the total resistance is maximum, which happens when rheostat resistance R=30Ω.


So the current will be,



Where Rtot is



Hence the minimum current, imin is



Similarly, maximum current, imax can be obtained when Rheostat resistance R is minimum, R=0Ω.


So, Rtot is



Hence the current is,



Hence the current in the ammeter vary from 0.15A to 0.83A


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