Answer :
Final answer:
The mutual core flux in the transformer is calculated to be 51.7 mWb. The primary and secondary leakage reactances are then found to be 3.24 Ω and 0.324 Ω respectively. The total reactance can then be calculated by adding these two values together.
Explanation:
The mutual flux (Ф) in a transformer can be calculated using the equation Ф = V / (4.44fN). Here 'V' is the voltage (rms) of the winding, 'f' is the frequency of the AC supply, and 'N' is the number of turns in the winding. The total number of turns for the transformer is the 200 turns on the primary side (Np = 200). The frequency (f) is 50 Hz. And the voltage on the primary side when rated voltage is applied (Vp) is 2300 V. Substituting these values into the equation gives φ = 2300/(4.44*50*200) = 0.0517 Wb, or 51.7 mWb.
Next, we calculate the leakage reactance. The leakage flux is said to be 1% of the mutual flux calculated earlier. Therefore, the leakage flux is 0.01* 51.7 mWb = 0.517 mWb. The leakage reactance for each winding can then be calculated using the formula Xl = 2πfϕlN, where ϕl is the leakage flux and N is the number of turns in the winding. This can be calculated for both the primary and secondary windings. Xl1 = 2*π*50*0.000517*200 = 3.24 Ω for the primary winding, and Xl2 = 2*π*50*0.000517*20 = 0.324 Ω for the secondary winding. The total reactance referred to either side can then be calculated by taking the sum of these two reactances.
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