Question

In LC circuit the inductance L=40 mH and capacitance C=100 μF. If a voltage V(t)=10sin(314t) is applied to the circuit, the current in the circuit is given as:

Answer 1

I(t) = 0.5185 cos(314t) A

Answer 2

The problem asks for the current in an ideal LC circuit when an AC voltage source is applied.

1. Identify Given Parameters:

   • Inductance, L = 40 mH = 40 × 10⁻³ H = 0.04 H
   • Capacitance, C = 100 μF = 100 × 10⁻⁶ F = 10⁻⁴ F
   • Applied Voltage, V(t) = 10sin(314t) V

2. Extract Voltage Amplitude and Angular Frequency:

   From the given voltage equation V(t) = V₀sin(ωt), we can identify:

   • Peak voltage, V₀ = 10 V
   • Angular frequency, ω = 314 rad/s

3. Calculate Reactances:

   • Inductive Reactance (X_L): $X_L = \omega L$ $X_L = 314 \, \text{rad/s} \times 0.04 \, \text{H}$ $X_L = 12.56 \, \Omega$

   • Capacitive Reactance (X_C): $X_C = \frac{1}{\omega C}$ $X_C = \frac{1}{314 \, \text{rad/s} \times 10^{-4} \, \text{F}}$ $X_C = \frac{1}{0.0314} \, \Omega$ $X_C \approx 31.847 \, \Omega$

4. Determine Total Impedance (Z):

   For a series LC circuit, the impedance is the absolute difference between the inductive and capacitive reactances: $Z = |X_L - X_C|$ Since $X_C (31.847 \, \Omega) > X_L (12.56 \, \Omega)$, the circuit is capacitive. $Z = X_C - X_L$ $Z = 31.847 \, \Omega - 12.56 \, \Omega$ $Z = 19.287 \, \Omega$

5. Calculate Peak Current (I₀):

   The peak current is given by Ohm's law for AC circuits: $I_0 = \frac{V_0}{Z}$ $I_0 = \frac{10 \, \text{V}}{19.287 \, \Omega}$ $I_0 \approx 0.5185 \, \text{A}$

6. Determine the Phase Relationship:

   In a series LC circuit:

   • If $X_L > X_C$, the circuit is inductive, and current lags the voltage by 90° (π/2 radians).
   • If $X_C > X_L$, the circuit is capacitive, and current leads the voltage by 90° (π/2 radians).
   • If $X_L = X_C$, the circuit is at resonance, and current is in phase with voltage (ideally infinite current in an ideal LC circuit).

   In this case, $X_C > X_L$, so the circuit is capacitive, and the current leads the voltage by π/2 radians. Since the applied voltage is $V(t) = V_0\sin(\omega t)$, the current will be $I(t) = I_0\sin(\omega t + \frac{\pi}{2})$. Using the trigonometric identity $\sin(x + \frac{\pi}{2}) = \cos(x)$, we get: $I(t) = I_0\cos(\omega t)$

7. Write the Current Equation:

   Substitute the calculated values of $I_0$ and $\omega$: $I(t) = 0.5185 \cos(314t) \, \text{A}$

The final answer is $\boxed{I(t) = 0.5185 \cos(314t) \, \text{A}}$.

Explanation of the solution:

1. Calculate inductive reactance ($X_L = \omega L$) and capacitive reactance ($X_C = 1/(\omega C)$) using the given L, C, and ω from the voltage equation.
2. Determine the total impedance ($Z = |X_L - X_C|$). Since $X_C > X_L$, the circuit is capacitive.
3. Calculate the peak current ($I_0 = V_0/Z$).
4. Since the circuit is capacitive, the current leads the voltage by $\pi/2$. Given $V(t) = V_0\sin(\omega t)$, the current will be $I(t) = I_0\sin(\omega t + \pi/2) = I_0\cos(\omega t)$.

Answer: The current in the circuit is given as: $I(t) = 0.5185 \cos(314t) \, \text{A}$