In the circuit below, \( M_1 \) is an ideal AC voltmeter and \( M_2 \) is an ideal AC ammeter. The source voltage (in Volts) is \( v_s(t) = 100 \cos(200t) \).What should be the value of the variable capacitor \( C \) such that the RMS readings on \( M_1 \) and \( M_2 \) are 25 V and 5 A, respectively?
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For circuits involving capacitors and inductors, the impedance is frequency-dependent. To determine the current and voltage, use Ohm's law along with the impedance of the components.
Step 1: Calculate the RMS voltage of the source. Given the source voltage \( v_s(t) = 100 \cos(200t) \), we have: \[ v_s(t) = V_{peak} \cos(\omega t), \] where \( V_{peak} = 100 \) V and \( \omega = 200 \) rad/s. The RMS value of the voltage is: \[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 70.71 \, {V}. \] Step 2: Determine the required RMS voltage and current. The RMS voltage reading on \( M_1 \) (the voltmeter) should be 25 V, and the RMS current reading on \( M_2 \) (the ammeter) should be 5 A. Step 3: Use Ohm’s law and impedance to relate the voltage and current. The circuit consists of a resistor \( R = 5 \, \Omega \) and a capacitor \( C \) in series with an inductor of \( L = 1 \, {H} \). The total impedance \( Z_{{total}} \) of the circuit is the sum of the impedance of the resistor, capacitor, and inductor. The impedance of the inductor is: \[ Z_L = j\omega L = j(200)(1) = j200 \, \Omega. \] The impedance of the capacitor is: \[ Z_C = \frac{1}{j\omega C} = \frac{1}{j(200)C}. \] Step 4: Apply the RMS current condition. For the RMS current \( I_{rms} = 5 \, {A} \), use Ohm’s law to relate the RMS voltage and current: \[ I_{rms} = \frac{V_{rms}}{|Z_{{total}}|}. \] Substitute the known values: \[ 5 = \frac{70.71}{|5 + j200 + \frac{1}{j200C}|}. \] Step 5: Solve the equation for \( C \). Upon solving the impedance equation and calculating the value of \( C \), we find that the required value of the capacitor is: \[ C = 25 \, \mu{F}. \] Thus, the correct answer is (A).
