What is the formula for the current in a series LC circuit?
2 Answers
In a series LC circuit, which consists of an inductor (L) and a capacitor (C) connected in series, the behavior of the current is influenced by the reactive properties of both components. The formula for the current \( I(t) \) in the circuit depends on the applied voltage and the initial conditions.
Here’s a step-by-step explanation of how to derive the formula for the current in a series LC circuit:
### 1. **Circuit Basics**
- **Inductor (L):** Has an inductive reactance \( X_L = \omega L \), where \( \omega = 2\pi f \) is the angular frequency of the AC source.
- **Capacitor (C):** Has a capacitive reactance \( X_C = \frac{1}{\omega C} \).
### 2. **Differential Equation**
In a series LC circuit, the voltage across the inductor \( V_L \) and the capacitor \( V_C \) must sum up to the total applied voltage \( V(t) \):
\[ V(t) = V_L + V_C \]
The voltage across each component can be expressed using their respective reactances:
- Voltage across the inductor: \( V_L = L \frac{dI(t)}{dt} \)
- Voltage across the capacitor: \( V_C = \frac{1}{C} \int I(t) \, dt \)
Combining these, we get:
\[ V(t) = L \frac{dI(t)}{dt} + \frac{1}{C} \int I(t) \, dt \]
To find the current \( I(t) \), we take the derivative of both sides with respect to time \( t \):
\[ \frac{dV(t)}{dt} = L \frac{d^2I(t)}{dt^2} + \frac{I(t)}{C} \]
Rearranging this gives the standard form of the differential equation for a series LC circuit:
\[ L \frac{d^2I(t)}{dt^2} + \frac{I(t)}{C} = \frac{dV(t)}{dt} \]
### 3. **Solution for Steady-State AC Voltage**
For an AC voltage source \( V(t) = V_0 \sin(\omega t) \), where \( V_0 \) is the amplitude and \( \omega \) is the angular frequency, the differential equation becomes:
\[ L \frac{d^2I(t)}{dt^2} + \frac{I(t)}{C} = V_0 \omega \cos(\omega t) \]
The particular solution for this differential equation is:
\[ I(t) = I_0 \sin(\omega t + \phi) \]
where \( I_0 \) is the amplitude of the current and \( \phi \) is the phase angle.
To find \( I_0 \) and \( \phi \), use the impedance of the circuit:
- The impedance \( Z \) of the LC circuit is \( \sqrt{ (X_L - X_C)^2 } \) which simplifies to \( |X_L - X_C| \) since the reactances are of opposite signs.
- For an LC circuit where \( X_L = X_C \), the impedance is zero, which implies resonance.
### 4. **Impedance and Current**
The impedance \( Z \) of a series LC circuit is:
\[ Z = \sqrt{ (X_L - X_C)^2 } = \sqrt{ (\omega L - \frac{1}{\omega C})^2 } \]
So,
\[ I_0 = \frac{V_0}{Z} = \frac{V_0}{| \omega L - \frac{1}{\omega C} |} \]
Thus, the current in the circuit is:
\[ I(t) = \frac{V_0}{| \omega L - \frac{1}{\omega C} |} \sin(\omega t + \phi) \]
where \( \phi \) is the phase angle determined by the relationship between \( X_L \) and \( X_C \).
### Summary
The formula for the current \( I(t) \) in a series LC circuit with an AC source \( V(t) = V_0 \sin(\omega t) \) is:
\[ I(t) = \frac{V_0}{| \omega L - \frac{1}{\omega C} |} \sin(\omega t + \phi) \]
where \( \phi \) is the phase angle, which depends on the initial conditions and the relative magnitudes of \( \omega L \) and \( \frac{1}{\omega C} \).
Here’s a step-by-step explanation of how to derive the formula for the current in a series LC circuit:
### 1. **Circuit Basics**
- **Inductor (L):** Has an inductive reactance \( X_L = \omega L \), where \( \omega = 2\pi f \) is the angular frequency of the AC source.
- **Capacitor (C):** Has a capacitive reactance \( X_C = \frac{1}{\omega C} \).
### 2. **Differential Equation**
In a series LC circuit, the voltage across the inductor \( V_L \) and the capacitor \( V_C \) must sum up to the total applied voltage \( V(t) \):
\[ V(t) = V_L + V_C \]
The voltage across each component can be expressed using their respective reactances:
- Voltage across the inductor: \( V_L = L \frac{dI(t)}{dt} \)
- Voltage across the capacitor: \( V_C = \frac{1}{C} \int I(t) \, dt \)
Combining these, we get:
\[ V(t) = L \frac{dI(t)}{dt} + \frac{1}{C} \int I(t) \, dt \]
To find the current \( I(t) \), we take the derivative of both sides with respect to time \( t \):
\[ \frac{dV(t)}{dt} = L \frac{d^2I(t)}{dt^2} + \frac{I(t)}{C} \]
Rearranging this gives the standard form of the differential equation for a series LC circuit:
\[ L \frac{d^2I(t)}{dt^2} + \frac{I(t)}{C} = \frac{dV(t)}{dt} \]
### 3. **Solution for Steady-State AC Voltage**
For an AC voltage source \( V(t) = V_0 \sin(\omega t) \), where \( V_0 \) is the amplitude and \( \omega \) is the angular frequency, the differential equation becomes:
\[ L \frac{d^2I(t)}{dt^2} + \frac{I(t)}{C} = V_0 \omega \cos(\omega t) \]
The particular solution for this differential equation is:
\[ I(t) = I_0 \sin(\omega t + \phi) \]
where \( I_0 \) is the amplitude of the current and \( \phi \) is the phase angle.
To find \( I_0 \) and \( \phi \), use the impedance of the circuit:
- The impedance \( Z \) of the LC circuit is \( \sqrt{ (X_L - X_C)^2 } \) which simplifies to \( |X_L - X_C| \) since the reactances are of opposite signs.
- For an LC circuit where \( X_L = X_C \), the impedance is zero, which implies resonance.
### 4. **Impedance and Current**
The impedance \( Z \) of a series LC circuit is:
\[ Z = \sqrt{ (X_L - X_C)^2 } = \sqrt{ (\omega L - \frac{1}{\omega C})^2 } \]
So,
\[ I_0 = \frac{V_0}{Z} = \frac{V_0}{| \omega L - \frac{1}{\omega C} |} \]
Thus, the current in the circuit is:
\[ I(t) = \frac{V_0}{| \omega L - \frac{1}{\omega C} |} \sin(\omega t + \phi) \]
where \( \phi \) is the phase angle determined by the relationship between \( X_L \) and \( X_C \).
### Summary
The formula for the current \( I(t) \) in a series LC circuit with an AC source \( V(t) = V_0 \sin(\omega t) \) is:
\[ I(t) = \frac{V_0}{| \omega L - \frac{1}{\omega C} |} \sin(\omega t + \phi) \]
where \( \phi \) is the phase angle, which depends on the initial conditions and the relative magnitudes of \( \omega L \) and \( \frac{1}{\omega C} \).
In a series LC circuit, the formula for the current \( I(t) \) as a function of time depends on the initial conditions and the applied voltage. Here's a detailed explanation:
### 1. **Series LC Circuit Basics**
A series LC circuit consists of an inductor (L) and a capacitor (C) connected in series with an alternating current (AC) source. The voltage across the circuit is \( V(t) \), and the circuit's total impedance is the combined impedance of the inductor and capacitor.
### 2. **Impedance of LC Circuit**
The impedance \( Z \) of the LC circuit is given by:
\[ Z = j \omega L - \frac{j}{\omega C} \]
where:
- \( j \) is the imaginary unit,
- \( \omega \) is the angular frequency of the AC source (in radians per second),
- \( L \) is the inductance of the inductor,
- \( C \) is the capacitance of the capacitor.
The impedance simplifies to:
\[ Z = j \left( \omega L - \frac{1}{\omega C} \right) \]
### 3. **Current Calculation**
To find the current \( I(t) \), use Ohm's Law for AC circuits:
\[ I(t) = \frac{V(t)}{Z} \]
Assuming a sinusoidal voltage source \( V(t) = V_0 \sin(\omega t) \), the current will also be sinusoidal and can be expressed as:
\[ I(t) = \frac{V_0 \sin(\omega t)}{j \left( \omega L - \frac{1}{\omega C} \right)} \]
### 4. **Simplified Form**
The impedance magnitude is:
\[ |Z| = \left| \omega L - \frac{1}{\omega C} \right| \]
The current magnitude is:
\[ I(t) = \frac{V_0}{|Z|} \sin(\omega t) \]
### 5. **Phase Relationship**
The phase angle \( \phi \) between the voltage and the current is given by:
\[ \phi = \arctan \left( \frac{\frac{1}{\omega C}}{\omega L} \right) \]
So, the current can be expressed as:
\[ I(t) = \frac{V_0}{|Z|} \sin(\omega t - \phi) \]
### 6. **Initial Conditions**
If the circuit is initially charged or if there's a DC component, the general solution might involve solving differential equations based on the initial charge and current. For a purely AC analysis, these initial conditions are typically not considered.
### Summary
In summary, for a sinusoidal AC voltage source \( V(t) = V_0 \sin(\omega t) \), the current in a series LC circuit is:
\[ I(t) = \frac{V_0}{\left| \omega L - \frac{1}{\omega C} \right|} \sin \left( \omega t - \arctan \left( \frac{\frac{1}{\omega C}}{\omega L} \right) \right) \]
This represents the time-varying current considering the impedance of the LC circuit.
### 1. **Series LC Circuit Basics**
A series LC circuit consists of an inductor (L) and a capacitor (C) connected in series with an alternating current (AC) source. The voltage across the circuit is \( V(t) \), and the circuit's total impedance is the combined impedance of the inductor and capacitor.
### 2. **Impedance of LC Circuit**
The impedance \( Z \) of the LC circuit is given by:
\[ Z = j \omega L - \frac{j}{\omega C} \]
where:
- \( j \) is the imaginary unit,
- \( \omega \) is the angular frequency of the AC source (in radians per second),
- \( L \) is the inductance of the inductor,
- \( C \) is the capacitance of the capacitor.
The impedance simplifies to:
\[ Z = j \left( \omega L - \frac{1}{\omega C} \right) \]
### 3. **Current Calculation**
To find the current \( I(t) \), use Ohm's Law for AC circuits:
\[ I(t) = \frac{V(t)}{Z} \]
Assuming a sinusoidal voltage source \( V(t) = V_0 \sin(\omega t) \), the current will also be sinusoidal and can be expressed as:
\[ I(t) = \frac{V_0 \sin(\omega t)}{j \left( \omega L - \frac{1}{\omega C} \right)} \]
### 4. **Simplified Form**
The impedance magnitude is:
\[ |Z| = \left| \omega L - \frac{1}{\omega C} \right| \]
The current magnitude is:
\[ I(t) = \frac{V_0}{|Z|} \sin(\omega t) \]
### 5. **Phase Relationship**
The phase angle \( \phi \) between the voltage and the current is given by:
\[ \phi = \arctan \left( \frac{\frac{1}{\omega C}}{\omega L} \right) \]
So, the current can be expressed as:
\[ I(t) = \frac{V_0}{|Z|} \sin(\omega t - \phi) \]
### 6. **Initial Conditions**
If the circuit is initially charged or if there's a DC component, the general solution might involve solving differential equations based on the initial charge and current. For a purely AC analysis, these initial conditions are typically not considered.
### Summary
In summary, for a sinusoidal AC voltage source \( V(t) = V_0 \sin(\omega t) \), the current in a series LC circuit is:
\[ I(t) = \frac{V_0}{\left| \omega L - \frac{1}{\omega C} \right|} \sin \left( \omega t - \arctan \left( \frac{\frac{1}{\omega C}}{\omega L} \right) \right) \]
This represents the time-varying current considering the impedance of the LC circuit.