Question

Calculate the following for the given circuit:
\includegraphics[]{5b phy fy.PNG} Given:
Resistance \( R = 400 \, \Omega \), Inductive voltage \( V_L = 120 \, V \), Capacitive voltage \( V_C = 120 \, V \), and source voltage: \[ V = 200 \cos(50\pi t) \, \text{volt} \] (i) Current in the circuit

Hint:

In an LC circuit, inductor voltage leads the current by \( 90^\circ \), and capacitor voltage lags by \( 90^\circ \), resulting in a \( 180^\circ \) phase difference.

Answer 1

Step 1: The impedance of the circuit: Since the inductor and capacitor voltages are equal, they cancel each other out, leaving only the resistance. Step 2: Applying Ohm's law: \[ I = \frac{V}{R} \] \[ = \frac{200}{400} \] \[ = 0.5 \, \text{A} \] \[ \boxed{I = 0.5 \, \text{A}} \] (ii) The potential across resistance % Solution Solution: Step 1: Using Ohm's law to find the potential across the resistor: \[ V_R = IR \] \[ = (0.5)(400) \] \[ = 200 \, \text{V} \] \[ \boxed{V_R = 200 \, \text{V}} \] (iii) The phase difference between the potentials across inductor and capacitor % Solution Solution: Step 1: In an LC circuit, the voltage across the inductor leads the current by \(90^\circ\), while the voltage across the capacitor lags the current by \(90^\circ\). Step 2: Therefore, the phase difference between the inductor and capacitor voltages is: \[ \theta = 90^\circ + 90^\circ = 180^\circ \] \[ \boxed{180^\circ} \]