HELP18

Question

Answer 1

2.1.1)

Q - factor (Quality factor): Q - factor of a resonant circuit is the measure of the "good ness" or "quality" of the circuit.

Q = X/R

Here, X is the capacitive reactance at resonance and R is the resistance .

Answer 2

2.1.2)

Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies.

It is typically measured in hertz , and may sometimes refer to pass band bandwidth, sometimes to base band bandwidth, depending on context.

Pass band bandwidth is the difference between the upper and lower cutoff frequencies of, for example, a band pass filter , a communication channel  or a signal spectrum .

In the case of a low - pass filter  or base band signal , the bandwidth is equal to its upper cutoff frequency.

Where BW is bandwidth,

             f₂ is upper frequency,

             f₁ is lower frequency.

Answer 3

(1)

Step 1:

Redraw the bridge network.

Find the current through the resistance .

Thevenins Theorem :

Thevenins Theorem states that it is possible to simplify any linear circuit, to an equivalent circuit with just a single voltage source and series resistance connected to a load.

Now reduce the circuit to a equivalent Thevenins circuit .

Find the Equivalent Resistance (Rs):

Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.

resistor is in series with the resistor.

resistor is in series with the resistor.

is in parallel with the .

Comments

Contd....

Step 2:

Find the Equivalent Voltage (Vs):

Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.

The potential is more easily calculated since the equivalent circuit reduces to two parallel voltage dividers. Using Ohm’s law, one find the current through the series arms, then the potential drop across the second resistor.

For example, the potential at point A is the current through this arm times the resistance of R₂.

Thus .

Simillarly, .

The differential potential is the potential of thevenin’s ideal potential source.

Step 3:

Therefore the Thevenins Equivalent circuit is shown below with the 10 Ω resistor connected.

The current across the load is

Solution :

The current across 10 Ω resistor is 16.7 mA.