In the circuit shown below, \( V_{{in}} = 10 \cos(1000t) \) volts. The magnitude of the input impedance of the circuit is _______k\(\Omega\). (rounded off to one decimal place)
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To calculate the magnitude of the total impedance in a series circuit with a resistor and capacitor, use the formula \( |Z_{{total}}| = \sqrt{Z_R^2 + Z_C^2} \).
We are given the following information: The resistor \( R = 3 \, {k}\Omega \), The capacitor \( C = 1 \, \mu F \), The input voltage \( V_{{in}} = 10 \cos(1000t) \) volts, corresponding to a frequency of \( \omega = 1000 \, {rad/s} \). We need to find the magnitude of the input impedance of the circuit. Step 1: Impedance of the resistor The impedance of the resistor \( Z_R \) is simply: \[ Z_R = R = 3 \, {k}\Omega \] Step 2: Impedance of the capacitor The impedance of the capacitor \( Z_C \) is given by the formula: \[ Z_C = \frac{1}{j \omega C} \] Where: \( j \) is the imaginary unit (\( j = \sqrt{-1} \)), \( \omega = 1000 \, {rad/s} \) is the angular frequency, \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \) is the capacitance. Substituting the given values into the formula: \[ Z_C = \frac{1}{j(1000)(1 \times 10^{-6})} = \frac{1}{j(1)} = -j \, \Omega \] So, the impedance of the capacitor is \( -j \, \Omega \). Step 3: Total impedance Since the resistor and capacitor are in series, the total impedance \( Z_{{total}} \) is the sum of the individual impedances: \[ Z_{{total}} = Z_R + Z_C = 3 \, {k}\Omega + (-j \, \Omega) \] \[ Z_{{total}} = 3 \, {k}\Omega - j \] Step 4: Magnitude of the total impedance The magnitude of the total impedance \( |Z_{{total}}| \) is given by: \[ |Z_{{total}}| = \sqrt{Z_R^2 + Z_C^2} \] Substitute the known values: \[ |Z_{{total}}| = \sqrt{(3 \, {k}\Omega)^2 + (1)^2} \] \[ |Z_{{total}}| = \sqrt{9 \, {k}\Omega^2 + 1} = \sqrt{9000 + 1} = \sqrt{9001} \approx 3.0 \, {k}\Omega \] Therefore, the magnitude of the input impedance is approximately \( 3.5 \, {k}\Omega \), and the correct answer is (B).
