Question

1. ( x_{lt} ) in this circuit is

a 1.8 kω

b 2.5 kω

c 6.5 kω

d 9.0 kω

8 khz

100 vpp

l1.13 h

l2.05 h

Explanation:

Step1: Calculate inductive reactance formula

The formula for inductive reactance is \(X_L = 2\pi fL\). For inductors in parallel, the total inductance \(L_{T}\) is given by \(\frac{1}{L_{T}}=\frac{1}{L_1}+\frac{1}{L_2}\). First, find \(L_{T}\):
\(\frac{1}{L_{T}}=\frac{1}{0.13}+\frac{1}{0.05}=\frac{0.05 + 0.13}{0.13\times0.05}=\frac{0.18}{0.0065}\), so \(L_{T}=\frac{0.13\times0.05}{0.13 + 0.05}=\frac{0.0065}{0.18}\approx0.0361\space H\)

Step2: Calculate total inductive reactance

Given \(f = 8\times10^{3}\space Hz\), use \(X_{LT}=2\pi fL_{T}\)
\(X_{LT}=2\times\pi\times8\times 10^{3}\times0.0361\)
\(X_{LT}=2\times3.14\times8\times10^{3}\times0.0361\)
\(X_{LT}=6.28\times8\times10^{3}\times0.0361\)
\(X_{LT}=50.24\times10^{3}\times0.0361\approx 1.8\times10^{3}\space\Omega = 1.8\space k\Omega\)

Answer:

A. \(1.8\space k\Omega\)