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CSecure I https/www.webassign.net/web/Student/Assignment-Responses/submit?dep-17748549 (a) A capacRor in ceries with a resistor, switch, and battery. (b) When the switch is thrown to position a. the capacitor begins to charpe up. ) When the cwitch is thrown to position b the capacitor discharges Evaluate the time constant of the crcuit = RC = (7.3 x 105 )(4.00 )-292 Evaluate the maximum charge on the capacitor: o-d8. (4.00 F)( 12.0 V) = (48 Evaluate the maximum current in the circuit: 12.0 V I, HA Find the charge and current as functions of time (in seconds): g(f) = ( HC) (Use the following as necessary.) HA) (Use the following as necessary: 1.) MASTER IT HINTS: EMSTUCK Now, a second resistor R2 of 3.6 × 105 is connected in parallel to the existing resistor in the circuit, and a second capacitor C2 = 4.50 F is connected to the existing capacitor m parallel. (a) What will be the new time constant? (b) What will be the maximum current in the dircuit (leaving the battery terminal)? Now, we connect the second resistor R2 of 3.6 x 10a and the second capacitor C,-4.50 pF both in series to the other elements of the circuit (c) What will be the new time constant? (d) what will be the maximum current in the circuit (leaving the battery terminal)? JA

Explanation / Answer

given

R2 = 3.6*10^5 ohm is connected in parallel to R1 = 7.3*10^5 ohm

hence

Reff = R1*R2/(R1 + R2) = 2.411*10^5 ohm

also, C2 = 4.5 uF

hence

Ceff = C1 + C2 = 8.5uF

a. hence new time constant = RC = 2.04935 s

b. maximum current in the circuit leaving the battery terminal = Io

Io = V/Reff = 4.9771 *10^-5 A

Io = 49.771 uA

c. for series connections

Reff = R1 + R2 = 10.9*10^5 ohm

Ceff = (C1C2)/(C1 + C2) = 2.1176 uF

hence

time constat, RC = 2.308235 s

Io = 12/Reff = 11 uA

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