QUESTION
The condition required to make the current through ZE = 0 is that when the impedances
SOLUTION
Zb =Zc and Za = Zd
And the voltages at node x and y are equal to each other. So it is the condition in which current through ZE = 0.
Ans. 2
(a) T to π conversion is as follows:
For pi conversion, we have the following formulae:
By putting values,
Pi Equivalent Circuit
(b) Thevenin equivalent of circuit:
For Thevenin equivalent of circuit, we short or turn off the independent voltage sources.
Vth is the open circuit voltage at the terminals and Rth is the input or equivalent resistance at the terminals.
We see from the output that Z1 and Z2 are in series so their equivalent will be
Now, Z3 is in parallel with Ztotal1.
So,
Ztotal2 is in series with Z4.
Ztotal3 is in parallel with Z5.
Ztotal4 is in series with Z6 and Z7.So,
This Zeq is equal to Zth.
So,
(c) For maximum power transfer, load should match each other. So, the value of load is
Where,
Ans 3. Data:
(a) Bandwidth:
Half-power frequencies:
At half upper and lower cut-off frequencies at -3dB are,
(b) If,
Then,
At resonance,
By Ohm’s law,
(c) The impedance offered by the circuit is
Impedance
Resonance occurs when voltage and current at the input terminals are in phase.
So,
So, the total impedance which offered to the supply circuit at resonance is the total impedance of the circuit which is purely resistive which is equal to the total value of
In parallel resonance circuit maximum impedance offered to supply.
Ans. 4 By applying Mesh analysis on the given circuit, we have 3 loops.
Loop 1:
I1 is passing through Z1 and Z3.
Loop 2:
I2 is passing through Z2 and Z4.
Loop 3:
I3 is passing through Z3 , Z4 and Z5.
So, the equations will become
…… (1)
…… (2)
…… (3)
From equation (1), we have
…… (4)
From equation (2), we have
…… (5)
By putting values of I1 and I2 in equation (3), we have
After putting the values and doing simplification, we have
To find I2, we put the value of I3 in equation (5).
By putting values, we have
To find I1, we put the value of I3 in equation (4).
By putting values, we have
Ans. 5 (a)
Kirchhoff’s equation to primary and secondary is
…… (1)
Resistance r1 and leakage reactance XL1 refer to the part of winding where only the primary current flows. Similarly, on the load side we have
By rearranging,
…… (2)
(b)
Where,
(c)
By Ohm’s law,
If L2 has 4 times as many windings then the step up was 4 times. So, input voltage is
Ans. 6 Data:
As it is lagging so,
% regulation
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