Dear Student When the ammeter is shunted, the resistance of the ammeteris, 1RA=1480+120=1+24480=25480or RA=48025=19 2 ohmWhen ammeter is connected with the battery and resistance in series then, RA+R=19 2+40 8=60 ohmSo current I = 3060=0 5A Regards

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Dear Student

W h e n t h e a m m e t e r i s s h u n t e d, t h e r e s i s \tan c e o f t h e a m m e t e r i s, \frac{1}{R_{A}} = \frac{1}{480} + \frac{1}{20} = \frac{1 + 24}{480} = \frac{25}{480} o r R_{A} = \frac{480}{25} = 19.2 o h m W h e n a m m e t e r i s c o n n e c t e d w i t h t h e b a t t e r y a n d r e s i s \tan c e i n s e r i e s t h e n, R_{A} + R = 19.2 + 40.8 = 60 o h m S o c u r r e n t I = \frac{30}{60} = 0.5 A

Regards

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