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
In the circuit shown in figure -...When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale defl...A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) hic...Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current $I = 10sin(\om...By error, a student places moving - coil voltmeter V (nearly ideal) in series with the resistance in a circuit in order ...