Question

For an ideal 2-winding transformer, the ampere-turns of the primary winding, ${\mathbf{N}}_{\mathbf{1}}{\mathbf{I}}_{\mathbf{1}}$, is equal to the of the secondary winding ${\mathbf{N}}_{\mathbf{2}}{\mathbf{I}}_{\mathbf{2}}$,

(a) True

(b) False

Short answer

Therefore, the correct option (a): True

Step-by-step solution

Find the relationship between MMF and core flux

The product of the turns and current in the magnetic circuit is directly proportional to the flux in the circuit.

$$\begin{gathered} \mathbf{NI}\mathbf{\propto }\mathbf{\varphi } \\ \mathbf{NI}\mathbf{=}\mathbf{\varphi }\mathfrak{R} \end{gathered}$$

Here,$N$ is the number of turns and $I$is the current, $\varphi$is the flus and $\mathfrak{R}$is the proportionality constant reluctance.

Find net MMF in the magnetic circuit.

Write the ohm’s law equation,

$N_1{I}_1-N_2{I}_2=\varphi \mathfrak{R}$…… (1)

The reluctance of the circuit is given by

$\mathfrak{R}=\frac{I_c}{{\mu }_c{A}_c}$

Here, $I_c$is the air gap length, ${\mu }_c$is the permeability of the medium and $A_c$is the cross-sectional area of the core.

The permeability of the ideal transformer is infinite. So, the reluctance of the ideal transformer becomes zero.

$$\begin{gathered} \mathfrak{R}=\frac{I_c}{\infty \times A_c} \\ \mathfrak{R}=0 \end{gathered}$$

From the equation (1)

$$\begin{gathered} N_1{I}_1-N_2{I}_2=\varphi \left(0\right) \\ {N}_1{I}_1-N_2{I}_2=0 \\ {N}_1{I}_1=N_2{I}_2 \end{gathered}$$

Hence, the ampere-turns of the primary winding is equal to the ampere-turn of the secondary winding.

Therefore, the correct option (a): True