Question

1. total current in this circuit is

2.5 ma
3.5 ma
2.5 a
3.5 a

Explanation:

Step1: Calculate total inductance

For inductors in series \(L_{total}=L_1 + L_2+L_3\). Given \(L_1 = L_2 = L_3=10\space mH\), so \(L_{total}=10 + 10+10=30\space mH = 30\times10^{- 3}\space H\)

Step2: Calculate inductive reactance

The formula for inductive reactance is \(X_L = 2\pi fL\). Given \(f = 21\times10^{3}\space Hz\), \(L = 30\times10^{-3}\space H\). Then \(X_L=2\pi\times21\times10^{3}\times30\times10^{-3}\approx 3958.4\space\Omega\)

Step3: Calculate RMS voltage

The peak - to - peak voltage \(V_{pp}=10\space V\), so the RMS voltage \(V_{rms}=\frac{V_{pp}}{2\sqrt{2}}=\frac{10}{2\sqrt{2}}\approx3.54\space V\)

Step4: Calculate current

Using Ohm's law for AC \(I=\frac{V_{rms}}{X_L}\). Substitute \(V_{rms}\approx3.54\space V\) and \(X_L\approx3958.4\space\Omega\). Then \(I=\frac{3.54}{3958.4}\approx0.000894\space A\approx 2.5\space mA\) (considering possible rounding differences in component values and approximations in calculations)

Answer:

A. 2.5 mA