Question

1. the voltage drop across l1 is________.a 2.3 vppb 14.8 vppc 15.0 vppd 17.1 vpp

Explanation:

Step1: Calculate inductive reactance \(X_L\)

The formula for inductive reactance is \(X_L = 2\pi fL\).
Given \(f = 10\times10^{3}\text{ Hz}\) and \(L=5\times 10^{- 3}\text{ H}\), then \(X_L=2\pi\times10\times10^{3}\times5\times10^{-3}=100\pi\Omega\approx314\Omega\)

Step2: Calculate the impedance \(Z\) of the series \(RL\) circuit

The impedance formula for a series \(RL\) circuit is \(Z=\sqrt{R^{2}+X_{L}^{2}}\). Given \(R = 2\times10^{3}\Omega\) and \(X_L\approx314\Omega\), then \(Z=\sqrt{(2000)^{2}+(314)^{2}}=\sqrt{4000000 + 98596}=\sqrt{4098596}\approx2024.5\Omega\)

Step3: Calculate the current \(I\) in the circuit

Using Ohm's law \(I=\frac{V}{Z}\), with \(V = 15\text{ V}_{PP}\) and \(Z\approx2024.5\Omega\), then \(I=\frac{15}{2024.5}\text{ A}\approx7.41\times10^{-3}\text{ A}\)

Step4: Calculate the voltage drop across the inductor \(V_L\)

The formula for the voltage drop across an inductor is \(V_L = IX_L\). Substituting \(I\approx7.41\times 10^{-3}\text{ A}\) and \(X_L\approx314\Omega\), we get \(V_L=(7.41\times10^{-3})\times314\approx2.3\text{ V}_{PP}\)

Answer:

A. \(2.3\text{ V}_{PP}\)