91Ó°ÊÓ

Primary turns are$N_p=500$.

Secondary turns are $N_s=10$.

A transformer is a device that transfers electrical energy from one alternating current circuit to one or more other circuits by either increasing (stepping up) or decreasing (stepping down) the voltage.

Ohm's law states that the current flowing through a conductor between two points is directly proportional to the voltage across both points.

Use the equation related to primary and secondary coil voltages and currents with the number of turns.

Atsecondary voltage with an open circuit:

$V_p=120V$

Using the equation as given below.

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

Rearranging it for $V_s$, you have

$$\begin{gathered} V_s=V_p\frac{N_s}{N_p} \\ =\left(120\right)\frac{10}{500} \\ =2.4V \end{gathered}$$

Hence, the secondary voltage at $V_p=120V$with an open circuit is 2.4 V .

Current through primary when secondary has resistive load$15\mathrm{Î\copyright }$.

Here,$l_s$is unknown, so we can find from Ohm’s law:

localid="1663226024205" $\frac{I_p}{I_s}=\frac{N_s}{N_p}$

Using the equation for current in transformer,

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

Rearranging it for primary current, we get

$$\begin{gathered} I_p=I_s\frac{N_s}{N_p} \\ =\left(0.16\right)\frac{10}{500} \\ =3.2×{10}^{-3}A \end{gathered}$$

Hence, the current through primary when secondary has resistive load $15\mathrm{Î\copyright }$is $3.2×{10}^{-3}A$.

Current through secondary when secondary has resistive load$15\mathrm{Î\copyright }$.

As current in secondary in the above part,

$I_s=0.16A$

Hence, the current through secondary when secondary has resistive load $15\mathrm{Î\copyright }$is 0.16 A.