Question

5) The initial voltages of 4 V and 24 V, respectively established by sources not shown (note polarity of the initial voltages). The switch is closed at t = 0. a. Find t) for t2 0; and i(t) for t 2 0 b. Find ) and ) for 2 0. Hint use tin part a and the current-voltage relationship for capacitors) c. Calculate the initial and final (that is, as t) energy stored in the capacitors. d. Calculate the total energy delivered to the 250 k2 resistor. capacitors C and C in the circuit below have been on t-0 C1 5 pF 250 kQ C2 24 V 20 μF Page 2 of 2 W-

Answer 1

At t = O^+, that is, the immediate instant after the switch is closed, [Also, t greaterthanorequalto 0] The circuit is: - ON solving the differential equation obtained using i(t) = i_c(t): where i_c(t) is the current across capacitors. For c_1 v_c(t) = v_c_1(infinity) + [V_c_1(0) - V_c_1(infinity)] e^-t/tau for C_2 V_c_2(t) = V_)c_2(infinity) + [V_c_2(c) - v_c_2(infinity)] e6-t/tau where V_c_1(infinity) and V_c_2(infinity) = 0 as the c_1 and c_2 will drain off Their energy after same time, rightarrow therefore v_c_1(t) = v_c_1(0) e^-t/tau = -4 e6-t/tau V_c_2(t) = v_c_2(0) e^-t/tau = 24 e6-t/tau v(t) = v_c_1(t) + v_c_2(t) = 20 e^-t/tau . . . .. .(1) rightarrow where tau = R_C_eq = (250 times 10^3) [5 times 10^-6 times 20 times 10^-6/25 times 10^-6] = 15 \r\n therefore using (1), v(t) = 20 e^-t/1 = 20 e^-t rightarrow Using I = dq/dt and q = CV therefore I(t) = C_eq d V(t)/dt = (4 times 10^-6) d/dt[20 e^-t] = 80