উত্তরঃ
To find the primary induced EMF (\(E_1\)), we use the phasor equation for the primary side of a transformer, considering the applied voltage, current, primary winding resistance, and leakage reactance.
Given values:
- Primary Resistance, \(R_1 = 0.15 \, \Omega\)
- Primary Leakage Reactance, \(X_1 = 0.8 \, \Omega\)
- Primary Current, \(I_1 = 60 \, A\)
- Power Factor, \(\cos \phi = 0.8\) lagging
- Rated Primary Voltage, \(V_1 = 1000 \, V\) (from 1000/250 V transformer rating)
Step 1: Calculate the primary winding impedance (\(Z_1\)).
\[ Z_1 = R_1 + jX_1 = 0.15 + j0.8 \, \Omega \]
In polar form:
\[ |Z_1| = \sqrt{R_1^2 + X_1^2} = \sqrt{0.15^2 + 0.8^2} = \sqrt{0.0225 + 0.64} = \sqrt{0.6625} \approx 0.8139 \, \Omega \]
\[ \theta_{Z_1} = \arctan\left(\frac{X_1}{R_1}\right) = \arctan\left(\frac{0.8}{0.15}\right) = \arctan(5.333) \approx 79.38^\circ \]
\[ Z_1 \approx 0.8139 \angle 79.38^\circ \, \Omega \]
Step 2: Determine the phase angle of the current.
Given \(\cos \phi = 0.8\).
\[ \phi = \arccos(0.8) \approx 36.87^\circ \]
Since the power factor is lagging, the current lags the voltage. If we take the primary voltage \(V_1\) as the reference phasor, then the current \(I_1\) will be:
\[ V_1 = 1000 \angle 0^\circ \, V \]
\[ I_1 = 60 \angle -36.87^\circ \, A \]
Step 3: Calculate the voltage drop across the primary impedance (\(I_1Z_1\)).
\[ I_1Z_1 = (60 \angle -36.87^\circ) \times (0.8139 \angle 79.38^\circ) \]
\[ |I_1Z_1| = 60 \times 0.8139 \approx 48.834 \, V \]
\[ \theta_{I_1Z_1} = -36.87^\circ + 79.38^\circ = 42.51^\circ \]
\[ I_1Z_1 \approx 48.834 \angle 42.51^\circ \, V \]
Convert \(I_1Z_1\) to rectangular form:
\[ (I_1Z_1)_{\text{real}} = 48.834 \cos(42.51^\circ) \approx 48.834 \times 0.737 = 36.00 \, V \]
\[ (I_1Z_1)_{\text{imag}} = 48.834 \sin(42.51^\circ) \approx 48.834 \times 0.675 = 33.00 \, V \]
\[ I_1Z_1 \approx 36.00 + j33.00 \, V \]
Step 4: Calculate the primary induced EMF (\(E_1\)).
The phasor equation for the primary circuit is:
\[ V_1 = E_1 + I_1Z_1 \]
Therefore,
\[ E_1 = V_1 - I_1Z_1 \]
\[ E_1 = (1000 + j0) - (36.00 + j33.00) \]
\[ E_1 = (1000 - 36.00) + j(0 - 33.00) \]
\[ E_1 = 964.00 - j33.00 \, V \]
Step 5: Find the magnitude of \(E_1\).
\[ |E_1| = \sqrt{964.00^2 + (-33.00)^2} \]
\[ |E_1| = \sqrt{929296 + 1089} \]
\[ |E_1| = \sqrt{930385} \approx 964.56 \, V \]
The primary induced electromotive force (EMF) in a transformer is the voltage induced in the primary winding due to the changing magnetic flux. It effectively opposes the applied primary voltage, \(V_1\). Its magnitude is determined by subtracting the voltage drop across the primary winding's internal impedance (\(I_1Z_1\)) from the applied primary voltage. This impedance (\(Z_1\)), composed of resistance (\(R_1\)) and leakage reactance (\(X_1\)), accounts for energy losses and reactive power within the primary coil. The power factor of the current dictates the phase relationship between the current and voltage, which is crucial for accurate phasor calculations involving complex impedances.
