Consider the circuit in the figure below, in which V = 6 V, and the switch has b

Question

Consider the circuit in the figure below, in which V = 6 V, and the switch has been in position a for a long time. The switch is changed to position b at t = 0 s. What are the charge Q on the capacitor and the current I through the resistor at the times tabulated below?

(a) What is the time constant ?(tau) for this circuit? The answer has to be in units SECONDS

time Q ( Consider the circuit in the figure below, in which V = 6 V, and the switch has been in position a for a long time. The switch is changed to position b at t = 0 s. What are the charge Q on the capacitor and the current I through the resistor at the times tabulated below? (a) What is the time constant ?(tau) for this circuit? The answer has to be in units SECONDS

Explanation / Answer

a) Time constant = RC = 50 x 2 x 10-6 = 1 x 10^-4 sec

b) Voltage across capacitor after very long time will be = 6 V

Voltage across capacitor can not be changed simaltaneously

So , Using Q = CV   = 2 x 10^-6 x 6 = 12 x 10^-6 C   = 12 uC

And i = V / R = 6 / 50 = 0.12 A   or 120 mA

C) After that   Q   =   Q0* e^(-t / RC)

Q at t = RC   will be Q = Q0 * e^-1   = 12 uC /e = 4.41 uC .......Ans

Q = CV

V = 4.41 x 10^-6 / 2 x 10^-6 = 2.21 Volt

I = V/R = 2.21 / 50 = 0.0441 A or 44.15 mA ...........Ans

And d part u don't need , So here is your solution.

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