(A) Find the time constant of the circuit. Use the definition of the time consta

Question

(A) Find the time constant of the circuit. Use the definition of the time constant. RC-(8.00 x 10° )(5.00 × 10-6 F) = 4.00 s (B) Calculate the maximum charge on the capacitor. Apply Kirchhoff's loop rule to the RC circuit, going clockwise, which means that the voltage difference across the battery is positive and the differences across the cap negative acitor and resistor are (2) -9-1R =0 From the definition of capacitance and Ohm's law, we have AVcq/C and 4R =-IR. These are voltage drops, so they're negative. Also, AVbat- when the maximum charge q = Q is reached, I0. Solve Equation (2) for the maximum charge Q = (5.00 x 10-6 F)( 12.0 V) = 60.0 C Substitute to find the maximum charge (C) Find the charge on the capacitor after 6.00 s. Substitute into the time-charge equation: Q(1-e-t/r) = (60.0 C)(1-e-6.00 s/4.00 s) q = (D) Compute the potential difference across the resistor after 6.00 s Compute the voltage drop across the capacitor at that time. Solve Equation (1) for AVR and substitute =-9.32 V C 5.00 F AVR =-dhat-Av =-12.0 V-(-9.32 V) =-2.68 V (E) Find the current in the resistor after 6.00 s. Apply Ohm's law, using the results of part (d) (remember that&VR; -IR here) -Av -(-2.68 V) (8.00 x 109(2)= 3.35 x 10-6 A

Explanation / Answer

PRACTICE IT

a) T = RC = 7.5E5*4.8e-6 = 3.6 s

b) Q = CV = 13*4.8e-6 = 0.0000624 C

c) q = Q*e^-t/T = 0.0000624*[1- e^(-6.3/3.6)] = 0.0000516 C

d) V = 13*e^(-6.3/3.6) = 2.259 V

e) i = 2.259/7.5e5 = 3.01*10^-6 A

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