Question

1. the voltage drop across r1 is

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

Explanation:

Step1: Calculate inductive reactance \(X_L\)

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

Step2: Calculate total impedance \(Z\)

In a series \(RL\) circuit, \(Z=\sqrt{(R_1 + R_2)^2+X_L^2}\). Here \(R_1 = 1000\space\Omega\), \(R_2 = 1000\space\Omega\), so \(R_1+R_2 = 2000\space\Omega\). Then \(Z=\sqrt{2000^2 + 5000^2}=\sqrt{4\times10^{6}+25\times 10^{6}}=\sqrt{29\times10^{6}}\approx5385\space\Omega\)

Step3: Calculate current \(I\)

Using Ohm's law \(I=\frac{V}{Z}\). Given \(V = 120\space V_{pp}\), \(I=\frac{120}{5385}\approx0.0223\space A\)

Step4: Calculate voltage drop across \(R_1\)

Using \(V_{R1}=IR_1\). Substituting \(I = 0.0223\space A\) and \(R_1 = 1000\space\Omega\), we get \(V_{R1}=0.0223\times1000 = 22.3\approx22.4\space V_{pp}\)

Answer:

C. \(22.4\space V_{pp}\)