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
Create a free account to view solution
View Solution FreeTopic: Current Electricity and Electrical Instrument·Practice all Current Electricity and Electrical Instrument questions
More Current Electricity and Electrical Instrument Questions
The charge flowing through a resistance R varies with time t as Q=at - bt2 where a and b are positiveconstants. The tota...Find potential of J with : - respect to G -...During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm usin...In the circuit shown here, what is the value of the unknown resistor R so that the total resistance of the circuit betwe...A whetstone′s bridge is balanced with a resistance of 625 Ω in the third arm, where P, Q and S are in the 1st...