Current Electricity and Electrical InstrumentHard
Question
In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open, (q is charge on the capacitor and
is Capacitive time constant). Which of the following statement is correct ?

is Capacitive time constant). Which of the following statement is correct ?
Options
A.Work done by the battery is half of the energy dissipated in the resistor
B.At t=
, q = CV/2
, q = CV/2C.At t =
,q = CV(1-e-2)
,q = CV(1-e-2) D.At
, q = CV(1-e-1)
, q = CV(1-e-1)Solution
For charging of capacitor
q = q0 (1 - e- t/
) = CV (1 - e- t/
)
at t =
, q = CV(1 - e-2)
q = q0 (1 - e- t/
) = CV (1 - e- t/
)at t =
, q = CV(1 - e-2)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
A battery of internal resistance 4W is connected to the network of resistance as shown. In the order that the maximum po...A conducting square frame of side ′a′ and a long straight wire carrying current I are located in the same pl...Two identical batteries, each of emf 2V and internal resistance r = 1Ω are connected as shown. The maximum powe...An electric bell has a resistance of 5Ω and requires a current of 0.25 A to work it. Assuming that the resistance o...Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit v...