# For an ideal single-phase transformer with a primary-to-secondary turns ratio of N:1, the ratio of instantaneous input power to instantaneous output power is

For an ideal single-phase transformer with a primary-to-secondary turns ratio of N:1, the ratio of instantaneous input power to instantaneous output power is

### Right Answer is: 1 :1

#### SOLUTION

Concept:

Turns ratio for the transformer is given by

$\frac{{{V_p}}}{{{V_s}}} = \frac{{{I_s}}}{{{I_p}}} = \frac{{{N_p}}}{{{N_s}}}$

Where,

Vp is voltage on primary side

Vs is voltage on secondary side

Ip is current on primary side

Is is current on secondary side

Np is turns on primary side

Ns is turns on secondary side

Ratio of input power Pp to output power Ps

$\frac{{{P_p}}}{{{P_s}}} = \frac{{{V_p}{I_p}}}{{{V_s}{I_s}}}$

Calculation:

Given primary-to-secondary turns ratio

$\dfrac{{{N_P}}}{{{N_S}}} = \dfrac{N}{1}$

The instantaneous input power to instantaneous output power is given by

$\begin{array}{l}\dfrac{{{P_i}}}{{{P_o}}} = \dfrac{{{V_i}{I_i}}}{{(N{V_i})\left( {\dfrac{1}{N}{I_i}} \right)}}\\\\\dfrac{{{P_i}}}{{{P_o}}} = 1:1\end{array}$

Scroll to Top