The load circuit consists of an inductor and a resistor in series. The inductor has roughly the same characteristic as the static converter. The measured voltage is taken from across the resistors.ERMISCH,JOCHEN,DR.-ING.KUNCKEL,KARL-HEINZ,DR.-ING....
The equivalent circuit of a physical (real) inductor is represented by an inductor in series with a resistor and a capacitor in parallel. Fig 1. At audio frequencies we can assume that the parasitic capacitance will be insignificant and discard it for practical purposes. The new equivalent circu...
If the upper element is made controllable, then the regulator is known as a series regulator because the controlled element is in series with the load. If the lower leg of the divider is controlled, then the regulator is known as a shunt regulator because this element is shunted by the lo...
A series circuit consists of a resistor with R=20Ω, an inductor with L=1H, a capacitor with C=0.002\ F and a 12-V battery. If the initial charge and current are both 0, find the charge and current at time t. 相关知识点: 试题来源: 解析 Q(t)=((-e^(-10t))(250))(6co...
Now consider to have a resistive component R1 in series with the inductor L1. It is easy to investigate analytically that this resistor provides a damping impact and may suppress the resonant peak of the filter characteristic. As using an actual resistor causes losses, it can be virtually ...
The importance of the error voltage depends on its size relative to the true voltage being measured, which in turn is proportional to the cell impedance. Mutual-inductance errors appear in the measured EIS spectrum as an inductor of valueMin series with the cell’s impedance. ...
Variable Resistor/Rheostat 可变电阻(滑动变阻器) 电路的类型 Series, Parallel and Mixed Seriescircuits 串联电路 (connected end-to-end) Rtotal=∑iRiRtotal=∑iRi Parallelcircuits 并联电路 (both ends are connected) 1Rtotal=∑i1Ri1Rtotal=∑i1Ri ...
A simple example is a resistor in series with an LED. You would usually want to have a current limiting resistor in series with your LED so that you can control the amount of current through the LED. If too much current is going through your LED, it will burn out too fast. If too ...
4 Using a Boot Resistor The charge-pump circuit in Figure 1-1 uses CBOOT to boost the high-side gate supply above the supply voltage of the power stage. One method to reduce ringing is to include a boot resistor in series with the boot capacitor, which slows down the turn on speed ...
Note that the resistor R2 is in series with the parallel combination of R1, and RL so that the current I1, is given by I1=V2/[R2+(R1RL/R1+RL)]=12/[10+150/25)]=12/16=0.75A Using the current division technique we see that the current through the load resistor RL is given by...