Current Electricity and Electrical InstrumentHard
Question
When an ammeter of negligible internal resistance is inserted in series with circuit it reads 1A. When the voltmeter of very large resistance is connected across X it reads 1V. When the point A and are shorted by a conducting wire, the voltmeters measures 10 V across the battery. The internal resistance of the battery is equal to
Options
A.zero
B.0.5 Ω
C.0.2 Ω
D.0.1 Ω
Solution
= 1A ⇒ Vx = 1 = 1 × X ⇒ X = 1ΩWhen Y shorted I =
10 = 12 - Ir ⇒ 10 = 12 -
⇒ 10 + 10 r = 10 = 12 - 12r
⇒ 10r = 2 ⇒ r = 0.2Ω
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