In the following circuit, the switch S is closed at t = 0. The charge on the capacitor `C_(1)` as a function of time will be given by `(C_(eq)=(C_(1)C_(2))/(C_(1)+C_(2)))`
Consider the situation shown in figure.The switch is closed at t=0 when the capacitor C_(1) as a function of time t .
In the given circuit switch is closed at t = 0 . The charge flown in time t = T_c (where T_c is time constant).
The capacitor C_1 in the figure shown initially carries a charge q_0 . When the switches S_1 and S_2 are closed,capacitor C_1 is connected in series to a resistor R and a second capacitor C_2 which is initially uncharged. The charge flown through wires as a function of time is where C=(C_1C_2)/(C_1+C_2)
In the given circuit C_1 = C, C_2 = 2C, C_3 = 3C . If charge at the capacitor C_2 is Q. Then the charge at the capacitor C_3 will be
In the circuit shown below the switch is closed at t=0 . For 0 lt t lt R (C_(1)+C_(2)) , the current I_(1) in the capacitor C_(1) in terms of total current I is
Condsider . In the circuit shown is the switch can be shifted to positions 1 and 2 The charge on capacitor C_(1) when the switch is at position 1 is.
For given circuit charge on capacitor C, and C_(2) in steady state will be equal to
Consider two circuits given below. When switches S_(1) and S_(2) are closed at t = 0 , the charge on capacitors C_(1) and C_(2) change with time as shown in graph 1 and 2 respectively. The current i_(1) and i_(2) in the two circuits change as shown in graph 1' & 2' respectively. Write the variation of current in the second circuit as function of time after the switch is closed at t=0. e^(-1) =0.37
The circuit shows in Fig. is in the steady state with switch S_(1) closed. At t = 0,S_(1) is opened and switch S_(2) is closed. a. Derive expression for the charge on capacitor C_(2) as a function of time. b. Determine the first instant t , when the energy in the inductor becomes one-third of that in the capacitor.
In the circuit below, if dielectric is inserted inot C_(2) then the charge on C_(1) will
