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A 20 Ohm resistance is connected
across the secondary winding of a
single-phase power transformer
whose secondary voltage is 150 V.
Calculate the primary voltage if the
supply current is 5 A, neglecting
losses.


Sagot :

Answer: 225 V

Explanation:

Given:

Secondary voltage,V2 = 150v

Resistance is connected across secondary winding,∴R2 = 20Ω

Supply current, ie, I1 = 5A

[tex]\text{Now, secondary current,} \ $I_{2}=\frac{V_{2}}{R_{2}}=\frac{150}{20}=7.5$$\therefore I_{2}=7.5 \mathrm{~A}$For Transformers, we have $\frac{N_{2}}{N_{1}}=\frac{V_{2}}{V_{1}}=\frac{I_{1}}{I_{2}}$ so, Taking $\frac{N_{2}}{N_{1}}=\frac{l_{1}}{l_{2}}$ inserting all given \& obtained values, we get $\frac{N_{2}}{N_{1}}=\frac{5}{7.5}$$\therefore \text { Turns ratio }=\frac{N_{1}}{N_{2}}=\frac{7.5}{5}=\frac{3}{2}=3: 2$[/tex]

[tex]\text{Again taking,}\frac{N_{1}}{N_{2}}=\frac{V_{2}}{V_{1}}$ \\So, \\$V_{1}=\frac{N_{1}}{N_{2}} V_{2}\\$ inserting all values, we get, $V_{1}=\frac{7.5}{5} * 150=225$\\Therefore, \fbox{Primary voltage, {$V_{1}=225 \mathrm{~V}$}}[/tex]

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