Question

The resistance of the rheostat shown in figure is 30 12. Neglecting the ammeter resistance, the ratio of
minimum and maximum currents through the ammeter, as the rheostat is varied, will be:

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Answer 1

Answer:

The answer will be 149 :825

Explanation:

According to the problem  the rheostat is having the resistance of  30 12 ohm

Now we know when resistance is maximum current is minimum and vice versa

Now for maximum current , i(max)  at first we will find the r equivalent,

r(eq) = 20 x 10 /20+10 = 20/3 ohm

Therefore i(max) =  V/r = 5.5/20/3 = 16.5/20 A= 0.825 A

Now for minimum current, we will consider the rheostat

In this case the r(eq) = 20/3 + 30.12 = 110.36/3 ohm

Therefore i(min) = V/r = 5.5/110.36/3 = 0.149 A

Now the ratio of minimum and maximum currents i(min)/i(max)

                                                                              = 0.149 /0.825 = 149: 825