Take the last digit of your student number. 1f it is zero or one, use \"2\" inst

Question

Take the last digit of your student number. 1f it is zero or one, use "2" instead. This value is the emf of the battery in volts (assume 2 sf so "2" 2.0 Volts Show all work 5 HF 3900 (a) Calculate the time constant of this circuit (6 marks) 3900 (b) Calculate the charge in coulombs when the capacitor is T 6.4 uF fully charged. Use the value from your student number for the emf of the battery. (4marks) (c) If the effective capacitor, C, is now allowed to discharge 35 HF through the effective resistance, R, show that the expression 95 HF for the time taken for half of the initial charge to decay away IS: RCIn2 half (6 marks) (d) Calculate this time (2 marks) (e) Calculate the initial current, just as the equivalent capacitor starts to discharge. Assume that the capacitor was fully charged by the battery. (2 marks)

Explanation / Answer

a.) Equivalent resistance

= ( 390 * 390 ) / ( 390 + 390 )

= 195 ohms

Equivalent capacitance

1 / C_eq = 1 / 5 + 1 / ( 6.4 + 95 ) + 1 / 35

Ceq = 4.19 uF

time constant, tau = R * C

= 195 * 4.19

= 317.05 us ans

b. ) Q = C_eq * V

= 4.19 * 10 ^ -6 * 2

= 8.38 uC ans

c.) Q = Q_0 * e ^ ( - t / ( R * C ) )

Now Q = Q_0 / 2

1 / 2 = e ^ ( - T_half / ( R * C ) )

taking log on both side

T_half = R * C * ln( 2 ) ans

d.) T_half = 317.05 * ln ( 2 )

T_half = 219.76 us ans

e ) When capacitor is fully charged it acts like battery

current = V / Req

= 2 / 195

= 0.01 A ans

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