Question

1. the voltage drop across l1 is

a 111.3 vpp
b 48.8 vpp
c 22.4 vpp
d 12 vpp

Explanation:

Step1: Calculate the inductive reactance \(X_L\)

The formula for inductive reactance is \(X_L = 2\pi fL\).
Given \(f = 60\space Hz\) and \(L=13.2\space H\), then \(X_L=2\pi\times60\times13.2\approx 5000\space\Omega\)

Step2: Calculate the total impedance \(Z\)

The resistors \(R_1 = 1000\space\Omega\) and \(R_2 = 1000\space\Omega\). The total resistance \(R = R_1+R_2=2000\space\Omega\).
Using the formula for impedance in an \(RL\) - series circuit \(Z=\sqrt{R^{2}+X_{L}^{2}}\), substitute \(R = 2000\space\Omega\) and \(X_L\approx5000\space\Omega\)
\(Z=\sqrt{2000^{2}+5000^{2}}=\sqrt{4\times10^{6} + 25\times10^{6}}=\sqrt{29\times10^{6}}\approx5385\space\Omega\)

Step3: Calculate the current \(I\)

Using Ohm's law \(I=\frac{V}{Z}\), with \(V = 120\space V_{PP}\) and \(Z\approx5385\space\Omega\)
\(I=\frac{120}{5385}\approx0.0223\space A\)

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

Using \(V_L = IX_L\), substitute \(I\approx0.0223\space A\) and \(X_L\approx5000\space\Omega\)
\(V_L=0.0223\times5000 = 111.5\space V_{PP}\approx111.3\space V_{PP}\)

Answer:

A. \(111.3\space V_{PP}\)