Question

a plug - in transformer supplies 9.00 v to a video game system. (a) how many turns are in its secondary coil, if its input voltage is 120 v and the primary coil has 400 turns? (b) what is its input current when its output is 1.30 a?

Explanation:

Step1: Use the transformer equation

The transformer equation is $\frac{V_p}{V_s}=\frac{N_p}{N_s}$, where $V_p$ is the primary voltage, $V_s$ is the secondary voltage, $N_p$ is the number of turns in the primary coil, and $N_s$ is the number of turns in the secondary coil.
Given $V_p = 120\ V$, $V_s=9.00\ V$, and $N_p = 400$.
We can solve for $N_s$:
$N_s=\frac{V_sN_p}{V_p}$
Substitute the values:
$N_s=\frac{9.00\times400}{120}$

Step2: Calculate the value of \(N_s\)

$N_s=\frac{3600}{120}=30$

Step3: Use the power conservation (assuming ideal transformer \(P_p = P_s\))

Power formula \(P = IV\). So \(I_pV_p=I_sV_s\)
We want to find \(I_p\), and we know \(V_p = 120\ V\), \(V_s = 9.00\ V\), \(I_s=1.30\ A\)
$I_p=\frac{I_sV_s}{V_p}$
Substitute the values:
$I_p=\frac{1.30\times9.00}{120}$

Step4: Calculate the value of \(I_p\)

$I_p=\frac{11.7}{120}=0.0975\ A$

Answer:

a. The number of turns in the secondary coil is \(30\).
b. The input current is \(0.0975\ A\).