Published by Patrick Mutisya · 14 days ago
Recall and use the equation \$\frac{Vp}{Vs} = \frac{Np}{Ns}\$ where the subscripts p and s refer to the primary and secondary windings of a transformer.
A transformer consists of two (or more) coils of insulated wire wound on a common magnetic core. An alternating current (AC) in the primary coil creates a time‑varying magnetic flux in the core. This changing flux links the secondary coil and induces an emf according to Faraday’s law.
Suggested diagram: cross‑section of a transformer showing primary coil, secondary coil, magnetic core and direction of flux.
For an ideal transformer:
\$\frac{Vp}{Vs}= \frac{Np}{Ns}\$
| Type | Turns Ratio (\$Np : Ns\$) | Effect on \cdot oltage | Effect on Current |
|---|---|---|---|
| Step‑up | \$Np < Ns\$ | Secondary voltage \$Vs > Vp\$ | Secondary current \$Is < Ip\$ |
| Step‑down | \$Np > Ns\$ | Secondary voltage \$Vs < Vp\$ | Secondary current \$Is > Ip\$ |
Problem: A transformer has 500 turns on the primary and 200 turns on the secondary. The primary voltage is 240 V. Find the secondary voltage and the secondary current if the primary current is 0.5 A (assume an ideal transformer).
\$\frac{Np}{Ns}= \frac{500}{200}=2.5\$
\$\frac{Vp}{Vs}= \frac{Np}{Ns}\;\Rightarrow\;Vs = \frac{Vp Ns}{Np}= \frac{240\times200}{500}=96\text{ V}\$
\$Is = \frac{Vp Ip}{Vs}= \frac{240\times0.5}{96}=1.25\text{ A}\$
Thus \$Vs = 96\,\$V and \$Is = 1.25\,\$A.
The fundamental relationship for an ideal transformer is \$\frac{Vp}{Vs}= \frac{Np}{Ns}\$. This allows you to predict how the voltage and current will change between the primary and secondary based on the turns ratio. Remember that power is conserved (ignoring losses), so an increase in voltage comes with a decrease in current and vice‑versa.