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this can be done using 3 op-amps in series.inthe 1st one connect a capacitor for feedback with a key across its end to charge and discharge it.in 2nd cnect capacitor across 2-6 and across its ends connect batery with akey in sereis.in 3rd just connect r for feedbak.conect o/p of final to 1st opamp through R.take o/p across final 6and ground

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Why does an inductor offer high impedance to ac but very low impedance to dc?

Because an inductor resists a change in current. The equation of an inductor is ...di/dt = V/L... meaning that the rate of change of current is proportional to voltage and inversely proportional to inductance. Solve the differential equation in a sinusoidal forcing function and you get inductive reactance being ...XL = 2 pi f L


How reactance is equal to 2pifL?

The equation of an inductor is ...di/dt = L / V... which means that the rate of change of current is proportional to voltage and inversely proportional to inductance.Set this up in a sinusoidal forcing circuit and solve the differential equation, or use phasors, and you get ...XL = 2 pi f L... which means that inductive reactance is proportional to both frequency and inductance by the factor of 2 pi.I have not included the derivation, because I don't know how to do it, and it does not seem necessary. If someone wants to provide one, please feel free to refine the answer.


How do you solve combination circuits?

Generally,1. Convert parallel branches into series equivalents2. Solve for the total resistance3. Solve for individual voltages4. Solve for individual currents5. Solve for power


A 2.2 kohm resistor that dissipates 0.5 W what is the current?

Lets work the mathematics:Power equation, P = VIOhm's law, R = V/ICombine these equations to solve for current:P/I = VR = P/I2I2 = P/RI = (P/R)-2Substituting and solving:I = (0.5 / 2200)-2 = 15mAI = 15mA


How many watts will be dissipated by a 30 ohm resistor with 12 volts applied?

P = IV Where: P = power in watts, I = Current, and V= Voltage Using ohms law: V = IR where V=Voltage, I = Current, and R= Resistance First solve for I, I = V/R, 12/30 = .4 Then use the power equation: P = .4*12 = 4.8 Watts

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