LinkedIn Insight UD-TRG-BEE-Mumbai-Q4-Ans - Grad Plus


Que. 4(a) Using source transformation tech. calculate voltage across 4Ω

Solution :- Step :- 1
Convert 6V voltage source into current source

I=\frac VR=\frac63=2A

Step 2 :-
6 Ω & 3Ω resistors are connected in parallel


Step 3 :-
Convert 2 A current source to voltage source
V = IR = 2 ×2 = 4 V

Step 4 :-
2 Ω & 2 Ω resistors are connected in series
2 + 2 = 4 Ω

Step 5 :-
Convert 4 V voltage source to current source

I=\frac VR=\frac44=1A

Step 6 :-
1 A & 3 A current sources are connected in parallel.
∴ 1 + 3 = 4 A

Step 7 :-
Convert 4 A current source into voltage source


Convert\;I=\frac VR=\frac{16}{4+10+4}=\frac{16}{18}=0.89A
Voltage across 4 Ω
= IR = 0.89 × 4 = 3.56 V

Que. 4(b) Find the average and rms value of the waveform

Solution :-
i) \begin{array}{l}V_{av}=\int_0^T\frac{Vdt}T\\=\frac1T\int_0^2Vdt\\=\frac12\left(\int_0^1Vdt+\int_1^2Vdt\right)\end{array}


ii) VRMS

\begin{array}{l}=\sqrt{\frac12\left[{\left(\frac{t^3}3\right)^1}<em>0+{\left[t\right]^2}_1\right]}\\=\sqrt{\frac12\left[\left(\frac13-0\right)+\left(2-1\right)\right]}\\\\\end{array} \begin{array}{l}=\sqrt{\frac12\left[\frac13+1\right]}\\=\sqrt{\frac12\left(\frac43\right)}\\V</em>{RMS}=0.816V\\\\\end{array}

Que. 4(c) The power in a 3-Φ ckt in measured by two wattmeters. If the total power is 50 kW and Pf is 0.6 lagging. Find the reading of each wattmeter.
Solution :-
From the given condition ,
P1 + P2 = 50 kW ——- (1)
Cos Φ = 0.6 (lagging)
Φ = 53.13 º
????1− ????2 = 38.39 ——– (2)
Solving eqn. (1) & (2) simultaneously ,
????1=44.195 ????????

????2=5.805 ????????

Que. 4(d) Explain the working of centre tapped full wave rectifier .
Solution :-
In the case of centre-tap full wave rectifier, only two diodes are used, and are connected to the opposite ends of a centre-tapped secondary transformer as shown in the figure below. The centre-tap is usually considered as the ground point or the zero voltage reference point.

As shown in the figure, an ac input is applied to the primary coils of the transformer. This input makes the secondary ends P1 and P2 become positive and negative alternately. For the positive half of the ac signal, the secondary point D1 is positive, GND point will have zero volt and P2 will be negative. At this instant diode D1 will be forward biased and diode D2 will be reverse biased. the diode D1 will conduct and D2 will not conduct during during the positive half cycle. Thus, the positive half cycle appears across the load resistance RLOAD.

During the negative half cycle, the secondary ends P1 becomes negative and P2 becomes positive. At this instant, the diode D1 will be negative and D2 will be positive with the zero reference point being the ground, GND. Thus, the diode D2 will be forward biased and D1 will be reverse biased. The diode D2 will conduct and D1 will not conduct during the negative half cycle.

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