Electronics Engineering

Theæ definition of farad is as follows; SI unit of capacitance, formally defined to be the capacitance of a capacitor between the plates of whichæthere appears a potential difference of one volt when it is charged by a quantity of electricity equal to one coulomb.

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When the frequency is zero(i.e when dc power is supplied), capacitor is open is treated as open circuit having infinite resistance.

The voltage drop across the emitter-collector junction develops the output signal with the help of a resistor or two in series. The output is 'seen' at the collector.

We will always calculate rms value only since the average value of ac current or voltage is zero. So we are using rms values in the ac circuit to calculate the power and to solve an ac circuit.

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If current transformers are connected in parallel, they will superpose and supply a high current.

EDIT:

I suppose you drive the transformer with a voltage source.

By the way, it depends how you connect the transformers:

A - primary side in parallel and two independent loads at the output

B - primary sides in parallel and secondary sides in parallel

Say turns ratio is Np:Ns, then for each transformer you have

Vp/Np=Vs/Ns, Np*Ip=Ns*Is.

IE If Np=N, Ns=1, you have a voltage step down transformer

(Vs=Vp/n, Is=N*Ip)

A: You will impose the same secondary voltage to each load, that is Vs=Vp/N.

Input current is the sum of each output load multiplied by transformer turns ratio. If the two transformers have M and N turns ratio, you have Iin = M Is1 + N Is2 (Is1=Is2 only if loads are identical)

This architecture is used in modular / redundant systems.

B: Theoretically, it's like having a single transformer. In practice, I have never seen it. What you do is to split the current in two different cores, so that your losses will be lower for each core. That *could be handy if you wanna use ferrites which are undersized for your application and you don't want to buy a new one.

Connect the SCR in series with the armature or field and control the firing angle of the SCR which in turns determine the current flowing to the armature or field.

Breakdown depends on the electric field value in FETS (as in diodes and such, where you can find a junction).

Theoretically, you need to

- dope less the junction region of your device (like p-i-n diodes, the i (intrinsic) region is not doped in order to reduce E field peak, which occurs near the center of the device).

- raise the length of your device

Both of these two solutions will have the drawback of increasing your ON resistance.

At circuit design level, you can protect your devices with clamp diodes or something similar.

When base signal is given to transistor at the time of forward bais of emitter and collector the current flows from high potential to low potential

Clock is propagated from one T or JK flip flop to another hence it works.

A ripple counter works by the following principle. A clock pulse is applied to the first flip flop and the output of the first flip flop acts as the clock input to the second flip flop and the sequence continues in that order.

it is aa linear circuit that produces the output voltage from the fraction of its input. It is also known as potential divider.

power must be gotten thru moving parts to the coil

slip rings, festoons, tinsel

for stability

may be it is best for torsional misalignmnent

Zenor diode is allow to flow current in forward direction and it can also allow current in reverse direction when the voltage is above the certain value.

You will have to have some computer skills to learn how to do this. You have to put in the right code in order to get it to work.

the gold one did it

Only if you are a politician or priest.

The circuit resistance is likely to gradually drop and in such case it will cause the circuit to burn down.

When your ripple is a lot lower than the supply voltage, here is a rule of thumb:

for a single diode rectifier: U = I/(100*C) with I in ampere, C in uF

for example 1A supply, 1000 uF cap, gives Upp = 1/100e-3= 10 Volts.

Use 60 for 60 Hz line systems.

for a bridge diode retifier: U = I/(50*C) with I in ampere, C in uF

for example 1A supply, 1000 uF cap, gives Upp = 1/50e-3= 5 Volts peak to peak.

Use 120 for 60 Hz line systems.

Note: it is only an approximation, does not include resistances etc.

a properly designed motor wont overheat unless it is overloaded

this is something like 3/4" drill with a 1/4" shaft that is used to drill in steel

sometimes hole saws have a 1/4" mandrel that will hold the entire line of saws

the motor must be matched to the load