Question

A transformer has 500primary turns and $10\frac{1}{2}$ secondary turns. (a) If role="math" localid="1663224210090" ${\mathit{V}}_{\mathbf{p}}$is 120 V(RMS),what${\mathit{V}}_{\mathbf{s}}$ is with an open circuit? (b) If the secondary now has a resistive load ofrole="math" localid="1663224020026" $\mathbf{15}\mathbf{}\mathit{\Omega }$, what is the current in the primary? (c) If the secondary now has a resistive load of $\mathbf{15}\mathit{\Omega }$what is the current in the secondary?

Short answer

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

b. Current through primary ,when secondary has resistive load $15\Omega$is $3.2\times {10}^{-3}A$.

c. Current through secondary when secondary has resistive load $15\Omega$is $0.16A$.

Step-by-step solution

Listing the given quantities:

Primary turns are$N_p=500$.

Secondary turns are $N_s=10$.

Understanding the concept of Ohm’s law and transformer:

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.

(a) Calculation of the Secondary voltage at with an open circuit:

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 .

(b) Calculation of the Current through primary when secondary has resistive load 15Ω

Current through primary when secondary has resistive load$15\Omega$.

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

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

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

(c) Calculation of the current through secondary when secondary has resistive load 15Ω:

Current through secondary when secondary has resistive load$15\Omega$.

As current in secondary in the above part,

$I_s=0.16A$

Hence, the current through secondary when secondary has resistive load $15\Omega$is 0.16 A.