For the circuit of the following figure, sketch the total high frequency response on a Bode plot showing: The critical frequencies 12 V (a) (b) The roll-off rates Cc = 12 pF Che = 40 pF Ce = 8 pF 5.6 kQ 68 kN 0.47 µF Vo B = 120 0.47 µF 3.3 kN '10 k2 1.0 k2 , 20 μF
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- 6. The resistance X of an electrical component has a probability density function f(x) Ar(130 - ) for resistance values in the range 10R1 R5 In the above circuit, R1 = 102, R2 = 400, R3 = 30N, R4 = 202, R5 = 60N. The battery voltage is 12V. 1. The equivalent resistance of the circuit is = Ω 2. The source current is = A 3. The voltage drop across R3 is = V 4.The voltage drop across R2 is = V 5. The power dissipated across R1 is = W 12VConsider the circuit shown in Figure. 2.00 N 3.00Q 1.00 N 18.0 V 4.00 N 1. The equivalent resistance of the above circuit is = 2. The total current supplied by the battery is = A 3. The potential differences across 30, 10 is= 4. The Power dissipated in the 20 is = W & Power dissipated 42 is = W 5. The power dissipated across 10 is = W & Power dissipated 30 is = W7:35 2. In the figure below the resistors have resistances of R1 225 N, R2 = 475 N, and R3 = 650 N R1 + Battery R2 R: a) Write the equation for the equivalent resistance of the series circuit. b) Calculate the equivalent resistance of the series circuit.Q.4: answer one of the following A) Use mesh current method for the circuit in fig. 5, to Calculate the current in each branch B) For the: For the capacitors circuit shown in fig.6 find the Qr, Cr, and the total energy Wr . 20 Ω C1=3 µF 30 Q C2=1 µF 40 Ω C3=5 µF 60 2 50 2 10 2 20 V 100 V 50 V VT = 200V fig. 5 fig. 6102 C 3. For the circuit at right, find: 102 Paralel (a) The current at pts. A, B, C, D, & E. The power dissipated by each resistor. (c) The power supplied by the battery. (d) The potential differences VB- VA, Vc- VB, VD-Vc, VE-VA. 100 V EIn the circuit shown, R,=R2=R, and ɛ2 =. The power %3D 2 dissipated in R, is 40 W. Does battery &2 supply or absorb energy, what is the rate of energy &2 exchanges with the circuit? Assume that the batteries are ideal. R2=R R www www R R,=R www R Supply 48 J/s Supply 8 W O Absorb 8 W O Absorb 1 W O supply 120 J/s Absorb 12 W O Supply 1 W O Absorb 120 J/s O Supply 12 w Supply 4 J/s O None of them O Absorb 48 J/s Absorb 4 J/sA = 60 B = 70 C = 170 D = 7dv + 5v = 10B(t), dt v(0) = 0 %3D For this problem we are just going to look at the equation for the battery. A fully charged battery when connected to the circuit at time zero have a voltage source equation B(t) = 10e-0.5t %3D b. Our battery is off until one second then connects for one second, then is off for one second and does not recharge (it just retains the voltage it had when it was removed from the circuit) and then is turned on for one second, then it turned off for good. Give the function for the battery voltage in the circuit using the Heaviside function and the exponential function. B2(t) =For the circuit in the figure below, find the average power, the maximum instantaneous power, and the minimum instantaneous power delivered to the capacitor. Use f = 500 Hz, Vmax=20 V, and C = 1.4 μF. VAC (a) the average power W (b) the maximum instantaneous power w (c) the minimum instantaneous power wWhen the current in the portion of the circuit shown below is 2.40 A and increases at a rate of 0.420 A/s, the measured voltage is AVab = 8.85 V. When the current is 2.40 A and decreases at the rate of 0.420 A/s, the measured voltage is AVab = 4.90 V. Calculate the values of L and R. HINT: Use a Kirchhoff loop for each of the two scenarios and then simultaneously solve the two equations. Use the signs as given by Kirchhoff's Rules for the emf in the inductor. L = H R = Ω ll L R a 9,20 Auto Fit for: Data Set | Current | = A*exp(-C*t) A: 20.00 C: 0.04000 10 Correlation: 1.000 RMSE: 2.373E-16 amp Current (amp) 15 LO