Transformer Equivalent Circuit MCQ Quiz in தமிழ் - Objective Question with Answer for Transformer Equivalent Circuit - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 13, 2025
Latest Transformer Equivalent Circuit MCQ Objective Questions
Top Transformer Equivalent Circuit MCQ Objective Questions
Transformer Equivalent Circuit Question 1:
In the equivalent circuit of transformers the magnetizing circuit branch is drawn in parallel to supply voltage because
Answer (Detailed Solution Below)
Transformer Equivalent Circuit Question 1 Detailed Solution
The equivalent circuit of the Transformer:
I1 = Primary winding current
I2 ' =Secondary winding current referred to the primary side
I0=NO-load current
\(\overrightarrow {{I_1}} = \overrightarrow {{I_0}} + \overrightarrow {{I_2}'} \)
The no-load current has two components.
\(\begin{array}{l} \overrightarrow {{I_0}} = \overrightarrow {{I_w}} + \overrightarrow {{I_m}} \\ {I_O} = \sqrt {I_w^2 + I_m^2} \end{array}\)
Iw= Core loss component
Im=magnetizing component
- No-load current is almost the same when we apply the load also.
- The magnetizing branch circuit connected in parallel to the supply voltage as No-load current is constant while we apply load also.
- Therefore Core loss also constant as I0 remains constant.
Transformer Equivalent Circuit Question 2:
Iμ is ___________ Component of current in case of transformer.
Answer (Detailed Solution Below)
Transformer Equivalent Circuit Question 2 Detailed Solution
Concept:
When the AC power source is connected to a transformer, a current flows in the primary winding, even when the secondary winding is open-circuited. This is the current required to produce the flux in the ferromagnetic core.
This current is called exciting current (Ie)
It consists of two components:
Magnetization current \({I_\mu}\): The current required to produce the flux in the core of the transformer. if the magnitude of the magnetizing current is more its power factor will become low on the lagging side.
Core-loss current Ie: The current required to make up for the hysteresis and eddy current losses
\({I_\mu} = \sqrt {I_e^2 - I_c^2} \)
Where,
It is the magnetizing component of current
Ie is the excitation component of current
Ic core loss component of current
Transformer Equivalent Circuit Question 3:
A 110/220 V, 60 Hz, single phase, 1-kVA transformer has leakage reactance of 4%. Calculate its total leakage inductance (in mH) referred to 110 V side.
Answer (Detailed Solution Below) 1.25 - 1.35
Transformer Equivalent Circuit Question 3 Detailed Solution
Assuming the 110-V side 1 and the 220-V side to be side 2, N1/N2 = 0.5
\({V_{1\;rated}} = 110\;V,{I_{1\;rated}} = \frac{{1000}}{{110}} = 9.09\;A\)
\({Z_{1\;base}} = \frac{{{V_{1\;rated}}}}{{{I_{1\;rated}}}} = \frac{{110}}{{9.09}} = 12.1\;{\rm{\Omega }}\)
Total leakage reactance referred to side 1
XL= (p.u.) ( Zbase )
\(L= \frac{{0.04 \times {Z_{1\;base}}}}{{2\pi \times 60}}\)
= 1.28 mHTransformer Equivalent Circuit Question 4:
A 50 Hz, single phase transformer draws a short circuit current of 30 A at 0.2 power factor lag when connected to 16V, 50 Hz source. The short circuit current when the same transformer is energized from 16V, 25 Hz source is____ (in A)
Answer (Detailed Solution Below) 56 - 57
Transformer Equivalent Circuit Question 4 Detailed Solution
\(\begin{array}{l} \frac{{{r_e}}}{{{z_e}}} = \frac{{{r_e}}}{{\sqrt {r_e^2 + x_e^2} }} = 0.2\\ 24r_e^2 = x_e^2\\ 30 = \frac{{16}}{{\sqrt {r_e^2 + x_e^2} }} = \frac{{16}}{{5{r_e}}}\\ \Rightarrow {r_e} = 0.1067\;{\rm{\Omega }}\\ {x_e} = 0.523\;{\rm{\Omega }} \end{array}\)
At 25 Hz, \({x_e} = 0.2615\;{\rm{\Omega }},\;{r_e} = 0.1067\;{\rm{\Omega }} \Rightarrow {z_e} = 0.2822\;{\rm{\Omega }}\)
\( \Rightarrow I = \frac{V}{{{z_e}}} = \frac{{16}}{{0.2822}} = 56.65\;A\)