Generate Quiz 6A357A750C77F

NPN transistors allow current to move from the emitter to the collector while controlling the current flow through the base. IE is therefore equal to IC+IB.

The AC power input =IRMS2(RF+Rr).
IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA
So, AC power INPUT=(13.5*10-3)2*(1500+10)=275.2mW

For instance, half-wave rectifying an input alternating voltage of 2 Hz (2 cycles per second) will result in an output signal with two peaks in a single second, one from each cycle. The output will therefore have 50 pulses or peaks if a 50 Hz alternating voltage is half-wave rectified. Option B is therefore accurate.

Step-up transformers lower the current while raising the voltage. Therefore, the response is D.

Step-down transformers are denoted by B.

The period is the amount of time needed to finish one full cycle.

B is accurate since the two microfarads and four microfarads capacitors will receive the same voltage when connected in parallel. However, because C is inversely proportional to voltage when the charge is equal, the four microfarads capacitor will receive a lower voltage because its capacitance is greater. As a result, it will be 1/2CV^2, meaning that the 4 microfarads capacitor will store more energy.

The 14.14V RMS voltage is represented by this option. Given the specified peak voltage of 20V, this is the right response. The peak voltage is divided by the square root of two to determine the RMS voltage.

The following formula can be used to determine the RMS (Root Mean Square) voltage across a full-wave rectifier:

Vrms = Vpeak / √2

We can determine the RMS voltage using the following formula since the full-wave rectifier’s peak voltage is 20V:

Vrms = 20V / √2

VRMs ≈ 14.14V