# UD-TRG-BEE-Mumbai-Q5-Ans

Que. 5(a) Find current through 30 Ω using thevenisis theorams.

Solution :-Unsolved

Que. 5(b) For the shown ckt. Find supply current in each branch & total Pf.

Solution :-

Data :- R1 = 10 Ω
R2 = 20 Ω
L1 = 0.12 H
C1 = 40 μF
V = 200 V
f = 50 Hz

\begin{array}{l}X_{L1}=2\pi fL\times2\times\pi\times50\times0.12=37.7\Omega\\X_{C1}=\frac1{2\pi fc}=\frac1{2\pi\times50\times40\times10^{-6}}=79.58\Omega\\\\\\\\\\\\end{array}

Z1 = R + jX????1=10+????.37.7=39 ∠ 75.14 Ω
Z2 = R – jX????1=20−????.79.58=82.05 ∠−75.9 ????
ZT = Z1 + Z2 = 51.51 ∠ – 54.39 Ω

\begin{array}{l}I_1=\frac V{Z_1}=\frac{200}{39<75.14}=5.13<-75.14A\\I_2=\frac V{Z_2}=\frac{200}{82.05<-75.9}=2.44<75.9A\\\\\\\\\\\\\\end{array} \begin{array}{l}I=\frac V{Z_T}=\frac{200}{51.51<-54.39}=3.88<54.39A\\\\\\\\\\\\\\end{array}

Total pf = cos????????
= cos(−54.39)
= 0.582 º

Que. 5(c) A 1000/200 V , 50 Hz , 1-phase transformer gave the following test results
OC test ( hv side ) : 100 V 0.24 A 90 W
SC test ( hv side ) : 50 V 5 A 110 W
Draw equivalent ckt of transformer reffered to primary and secondary side.
Solution :-
Data :- a) OC test
Pi = 90 W
I0 = 0.24 A
V1 = 1000 V
Pi = V1I0cosΦ0

\begin{array}{l}\Phi_0=\cos^{-1}\left(\frac{P_i}{V_1I_0}\right)\\=\cos^{-1}\left(\frac{90}{1000\times0.24}\right)\\=67.975^{0w}\\\\\\\\\\\\\\end{array}

Pi = V1Iw
\begin{array}{l}I_w=\frac{P_i}{V_1}=0.09A\\I_{0^2}=I_{w^2}+I_{u^2}\\\\\\\\\\\\\\end{array}
\begin{array}{l}I_u=\sqrt{{I^2}_0-{I^2}_w}\\=\sqrt{0.24^2-0.09^2}\\=0.22A\\\\\\\\\\\\\\\\end{array}
\begin{array}{l}I_wI_0=V_1\\R_0=\frac{V_1}{I_w}\\=\frac{1000}{0.09}\\=11111.11\Omega\\\\\\\\\\\\\\\\\\end{array}
\begin{array}{l}I_wI_0=V_1\\R_0=11.111K\Omega\\I_uX_0=V_1\\X_0=\frac{V_1}{I_u}\\\\\\\\\\\\\\\\\\end{array}

\begin{array}{l}I_wI_0=V_1\\=\frac{1000}{0.22}\\=4545.45\Omega\\X_0=4.545K\Omega\\\\\\\\\\\\\\\\\\end{array}

b) SC test
Pcfl = 110 w
I1fl = 5 A
VSC = 50 V
Equivalent ckt constant to primary

\begin{array}{l}I_wI_0=V_1\\{I^2}<em>{1fl}R</em>{e1}=P_{cfl}\\R_{e1}=\frac{P_{cfl}}{{I^2}<em>{1fl}}=\frac{110}{5^2}=4.4\Omega\\V</em>{sc}=I_{1fl.Z_{e1}}\\\\\\\\\\\\\\\\\\end{array} \begin{array}{l}I_wI_0=V_1\\Z_{e1}=\frac{V_{SC}}{I_{1fl}}=\frac{50}5=10\Omega\\{R^2}<em>{e1}+{X^2}</em>{e1}={Z^2}<em>{e1}\\\\\\\\\\\\\\\\\\end{array} \begin{array}{l}I_wI_0=V_1\\X</em>{e1}=\sqrt{{Z^2}<em>{e1}-{R^2}</em>{e1}}\\=\sqrt{10^2-4.4^2}\\=8.98\Omega\\\\\\\\\\\\\\\\\\end{array} \begin{array}{l}I_wI_0=V_1\\\cos\Phi_{SC}=\frac{R_{e1}}{Z_{e1}}\\=\frac{4.4}{10}\\\cos\Phi_{SC}=0.44\\\\\\\\\\\\\\\\\\end{array} \begin{array}{l}\Phi_{SC}=\cos^{-1}\left(0.44\right)\\=63.98^0\\\\\\\\\\\\\\\\\\end{array}

Equivalent ckt constant referred to secondary
\begin{array}{l}Z_{e2}=Z_{e1}\left(\frac{V_2}{V_1}\right)^2=10\times\left(\frac{200}{1000}\right)^2\\=0.4\Omega\\\\\\\\\\\\\\\\\\\\end{array}
\begin{array}{l}R_{e2}=R_{e1}\left(\frac{V^2}{V^1}\right)^2=4.4\times\left(\frac{200}{1000}\right)^2\\=0.176\Omega\\\\\\\\\\\\\\\\\\\\end{array}

\begin{array}{l}X_{e2}=X_{e2}\left(\frac{V_2}{V_1}\right)^2=8.98\times\left(\frac{200}{1000}\right)^2\\=0.359\Omega\\\\\\\\\\\\\\\\\\\\end{array}

Equivalent circuit referred to primary side of transformer

Equivalent circuit referred to secondary side of transformer

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