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UD-TRG-BEE-Mumbai-Q3-Ans

Que. 3(a) A balanced three-phase load connected in delta , draws a power of 10 kW at 440 V at a Pf of 0.6 lead, find the values of circuit elements and reactive volt – ampere drawn .
Solution :-
For a delta connection
VL = VPh = 440 V
P = 10 kW
Pf = 0.6 lead
\begin{array}{l}P=\sqrt3V_LI_L\cos\Phi\\10\times10^3=\sqrt3\times440\times I_L\times0.8\\I_L=21.86A\\I_L=\sqrt3\times I_{Ph}\\\\\\\\end{array}

\begin{array}{l}Phase\;current\;I_{Ph}=\frac{I_L}{\sqrt3}=\frac{21.86}{\sqrt3}=12.621A\\\\\\\\end{array} \begin{array}{l}Phase\;impendenceZ_{Ph}=\frac{V_{Ph}}{I_{Ph}}=\frac{440}{12.621}=34.86\Omega\\Phase\;angle\Phi=\cos^{-1}\left(0.6\right)=53.13^0\end{array}

Resistance of each phase = RPh
= ZPh × cos Φ
= 34.86 × 0.6
= 20.92 Ω

Inductive reactance of each phase = XLPh
= ZPh × sin Φ
= 34.86 × 0.8
= 27.88 Ω

Reactive volt amperes

\begin{array}{l}Q=\sqrt3V_LI_L\sin\Phi=\sqrt3\times440\times21.86\times0.8\\=13328VAR\\=13.328KVAR\end{array}

Que. 3(b) The wattmeter reads losses in OC test and reads copper losses in SC test of a transformer. Justify .
Solution :-
The Open circuit & short circuit test are perform to determine the circuit contants , efficiency & regulation without actually loading the transformer.

Open – circuit test :-
It is usually high voltage winding – is left open and the other is connected to its supply of normal voltage and frequency. A wattmeter W, voltmeter V, and an ammeter A are connected in the low voltage winding.

With normal voltage applied to the primary, nomal flux will be set up in the core, hence normal iron losses will occur which are recorded by the wattmeter. As the primary no-load current IQ is small, Cu loss is negligibly small in primary and nil in secondary. Hence the wattmeter reading represent practically the core loss or iron loss under no-load condition.

Short – circuit test :-

In this test, usually the low-voltage winding, is solidly short circuited by a thick conductor. A low voltage at current frequency is applied to the primary and is cautiously increased till full load current are flowing both in primaryand secondary.

Since, in this test, the applied voltage is a small percentage of the normal voltage, the mutual flux Φ produced is also a small percentage of its normal value. Hence core losses or iron losses are very small with the result that the wattmeter reading represent the full-load Cu loss for the whole transformer.

Que. 3(c) What is meant by filter
Solution :-
A filter is a circuit which can be designed to modify , reshape or remove all the undesired frequencies of an electrical signals and pass only the desired signals .
It is usually a frequency selective network that passes a specified band of frequencies and block signals of frequencies outside this bands .

Que. 3(d) Draw and explain O/P characteristic of transistor in CE configuration
Solution :-
It is also referred as CE configuration . In Common emitter configuration , base is the input terminal , collector is the output terminal & emitter is the common terminal for both input & output.

Diagram

The common emitter amplifiers are used when large current gain is needed. The input signal is applied between the base and the emitter terminals while the output signals is taken between the collector and emitter terminals. Thus , the emitter terminal of the transistor is common for both input and output and hence it is named as Common emitter configuration .

O/P characteristics :-
• The output characteristics of a CE configuration are plot of collector to emitter voltage ( VCE ) to collector current IC for a constant base current IB.
• It has three regions ACTIVE , CUTOFF and SATURATION regions.

Active region :- In this region , the emitter-base junction is forward biased while the collector-base junction is reversed biased. When the value of VCE is zero , the collector current is negligibly small. For VCE between 0 and approximately 1V, the collector current rises sharply and becomes almost constant.

Saturation region :- In this region , the emitter-base and collector-base junction are forward biased. In this graph this region corresponds to portion where VCE < 0.2V .

Cutoff region :- In cutoff region , the emitter-base and collector-base
junctions of a transistor are reversed biased. Hence IC is not equal to zero when IB is zero.

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