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+1 vote
8.1k views
in Physics by (63.5k points)

A battery of unknown e.m.f is connected across resistances as shown in Fig. The voltage drop across the 8-Ω resistor is 20 V. What will be the current reading in the ammeter? What is the e.m.f of the battery?

1 Answer

+2 votes
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Best answer

Current through 8-Ω resistance =20/8 =2.5 A 

This current is divided into two parts at point A; one part going along path AC and the other along path ABC which has a resistance of 28Ω.

l2 = 2.5 x 11/(11 + 28) = 0.7

Hence, ammeter reads 0.7 A. 

Resistance between A and C =(28 x 11/39) ohm. 

Total circuit resistance = 8 + 11 + (308/39) =1049/39Ω

∴ E = 2.5 x 1049/39=67.3 V

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