In an ideal transformer, how is the secondary voltage related to turns?

In an ideal transformer, how is the secondary voltage related to turns?

• A. V_s = V_p
• B. V_s = (N_p/N_s) × V_p
• C. V_s = (N_p/N_s) × I_p
• D. V_s = (N_s/N_p) × V_p

Answer: (D)

Voltage on the secondary is determined by the turns on that winding because the induced emf in a coil is proportional to the number of turns when the magnetic flux in the core is the same for both windings. In an ideal transformer, both windings share the same core flux, so by Faraday’s law the voltages satisfy V_p / V_s = N_p / N_s. Rearranging gives V_s = (N_s / N_p) × V_p. This means the secondary voltage scales with the turns ratio: more turns on the secondary give a higher voltage, fewer turns give a lower voltage. If the turns are equal, voltages are equal; if the secondary has more turns, V_s is higher, and if it has fewer turns, V_s is lower. The other forms either invert this relationship or mix in current; in an ideal transformer the current relation V_p I_p = V_s I_s holds, but the voltage itself is set by the turns.

Voltage on the secondary is determined by the turns on that winding because the induced emf in a coil is proportional to the number of turns when the magnetic flux in the core is the same for both windings. In an ideal transformer, both windings share the same core flux, so by Faraday’s law the voltages satisfy V_p / V_s = N_p / N_s. Rearranging gives V_s = (N_s / N_p) × V_p. This means the secondary voltage scales with the turns ratio: more turns on the secondary give a higher voltage, fewer turns give a lower voltage. If the turns are equal, voltages are equal; if the secondary has more turns, V_s is higher, and if it has fewer turns, V_s is lower. The other forms either invert this relationship or mix in current; in an ideal transformer the current relation V_p I_p = V_s I_s holds, but the voltage itself is set by the turns.