Question

1. calculate ( i_t ).

a 6.67 ma
b 6.63 ma
c 9.4 ma
d 13.1 ma

Explanation:

Step1: Calculate inductive reactance \( X_L \)

The formula for inductive reactance is \( X_L = 2\pi fL \).
Given \( f = 100\ \text{Hz}\) and \( L = 12\ \text{H}\), then \( X_L=2\pi\times100\times12\approx 7539.82\ \Omega\).

Step2: Calculate currents through inductor \( I_L \) and resistor \( I_R \)

The peak - to - peak voltage \( V_{PP}=50\ \text{V}\), so the RMS voltage \( V=\frac{V_{PP}}{2\sqrt{2}}=\frac{50}{2\sqrt{2}}\approx17.68\ \text{V}\).
For the inductor, \( I_L=\frac{V}{X_L}\), substituting \( V = 17.68\ \text{V}\) and \( X_L\approx7539.82\ \Omega\), we get \( I_L=\frac{17.68}{7539.82}\approx2.358\times 10^{-3}\ \text{A}\).
For the resistor \( R = 7.5\times10^{3}\ \Omega\), \( I_R=\frac{V}{R}\), substituting \( V = 17.68\ \text{V}\) and \( R = 7.5\times 10^{3}\ \Omega\), we get \( I_R=\frac{17.68}{7.5\times 10^{3}}\approx2.357\times 10^{-3}\ \text{A}\).

Step3: Calculate total current \( I_T \)

In a parallel \( RL \) circuit, \( I_T=\sqrt{I_R^{2}+I_L^{2}}\).
\( I_T=\sqrt{(2.357\times 10^{-3})^{2}+(2.358\times 10^{-3})^{2}}\approx 3.33\times10^{-3}\ \text{A}\) (peak - to - peak current \( I_{T(PP)}=\sqrt{2}\times I_T\times2\)).
Another way (using peak - to - peak values directly):
The formula for inductive reactance \( X_L = 2\pi fL=2\pi\times100\times12\approx7539.82\ \Omega\)
Current through inductor \( I_{L(PP)}=\frac{V_{PP}}{X_L}=\frac{50}{7539.82}\approx6.63\times 10^{-3}\ \text{A}\)
Current through resistor \( I_{R(PP)}=\frac{V_{PP}}{R}=\frac{50}{7.5\times 10^{3}}\approx6.67\times 10^{-3}\ \text{A}\)
In a parallel \( RL \) circuit, \( I_T=\sqrt{I_{R(PP)}^{2}+I_{L(PP)}^{2}}\)
\( I_T=\sqrt{(6.67\times 10^{-3})^{2}+(6.63\times 10^{-3})^{2}}\approx9.4\times 10^{-3}\ \text{A} = 9.4\ \text{mA}\)

Answer:

C. 9.4 mA