The same current is flowing in two AC circuits. The first circuit contains a pure inductor and the second, a capacitor. If the frequency of the AC is increased, then the current will:
Decrease in the first and increase in the second
Analyzing Current in AC Circuits with Changing Frequency
This question asks about how the current changes in two different AC circuits – one with a pure inductor and one with a pure capacitor – when the frequency of the AC voltage source is increased, assuming the voltage remains constant.
In AC circuits, the components like inductors and capacitors oppose the flow of current. This opposition is called reactance. For an inductor, it's inductive reactance ($X_L$), and for a capacitor, it's capacitive reactance ($X_C$).
Inductive Reactance and Current
For a pure inductor of inductance $L$, the inductive reactance ($X_L$) is given by the formula:
\( X_L = \omega L \)
where \(\omega\) is the angular frequency of the AC source, and \(\omega = 2\pi f\), where $f$ is the frequency in Hertz. So, the formula can also be written as:
\( X_L = 2\pi f L \)
In a pure inductor circuit connected to an AC voltage source with voltage $V$, the current $I$ is given by Ohm's law for AC circuits:
\( I = \frac{V}{X_L} \)
Substituting the formula for $X_L$:
\( I = \frac{V}{2\pi f L} \)
From this formula, we can see that the current $I$ is inversely proportional to the frequency $f$ (assuming $V$ and $L$ are constant):
\( I \propto \frac{1}{f} \)
Therefore, if the frequency $f$ increases, the inductive reactance $X_L$ increases, and the current $I$ in the pure inductor circuit will decrease.
Capacitive Reactance and Current
For a pure capacitor of capacitance $C$, the capacitive reactance ($X_C$) is given by the formula:
\( X_C = \frac{1}{\omega C} \)
Substituting \(\omega = 2\pi f\), the formula becomes:
\( X_C = \frac{1}{2\pi f C} \)
In a pure capacitor circuit connected to an AC voltage source with voltage $V$, the current $I$ is given by Ohm's law for AC circuits:
\( I = \frac{V}{X_C} \)
Substituting the formula for $X_C$:
\( I = \frac{V}{\frac{1}{2\pi f C}} = V \cdot 2\pi f C \)
From this formula, we can see that the current $I$ is directly proportional to the frequency $f$ (assuming $V$ and $C$ are constant):
\( I \propto f \)
Therefore, if the frequency $f$ increases, the capacitive reactance $X_C$ decreases, and the current $I$ in the pure capacitor circuit will increase.
Summary of Frequency Effect on Current
Based on the analysis:
- In a pure inductor circuit, increasing frequency decreases the current.
- In a pure capacitor circuit, increasing frequency increases the current.
So, if the frequency of the AC is increased, the current will decrease in the first circuit (pure inductor) and increase in the second circuit (pure capacitor).
| Circuit Component | Reactance Formula | Relationship with Frequency (f) | Effect of Increasing Frequency on Reactance | Effect of Increasing Frequency on Current (I = V/Z) |
|---|---|---|---|---|
| Pure Inductor (L) | \(X_L = 2\pi f L\) | \(X_L \propto f\) | Increases | Decreases (\(I \propto 1/X_L\)) |
| Pure Capacitor (C) | \(X_C = \frac{1}{2\pi f C}\) | \(X_C \propto \frac{1}{f}\) | Decreases | Increases (\(I \propto 1/X_C\)) |
Conclusion
When the frequency of the AC source is increased, the current in the pure inductor circuit decreases, and the current in the pure capacitor circuit increases.
Revision Table: AC Circuit Concepts
| Concept | Description | Formula |
|---|---|---|
| Inductive Reactance | Opposition to current flow by an inductor in an AC circuit. | \(X_L = 2\pi f L\) |
| Capacitive Reactance | Opposition to current flow by a capacitor in an AC circuit. | \(X_C = \frac{1}{2\pi f C}\) |
| Ohm's Law (AC) | Relates voltage, current, and impedance (or reactance for pure components). | \(I = V/Z\) (where Z is impedance/reactance) |
Additional Information: AC Circuit Behavior
It's important to remember the phase relationship between voltage and current in these pure component circuits:
- In a pure inductor circuit, the voltage leads the current by 90 degrees (\(\pi/2\) radians). This means the current reaches its peak value a quarter cycle after the voltage reaches its peak.
- In a pure capacitor circuit, the current leads the voltage by 90 degrees (\(\pi/2\) radians). This means the current reaches its peak value a quarter cycle before the voltage reaches its peak.
- In a purely resistive AC circuit, the voltage and current are in phase.
These phase differences are crucial for understanding power in AC circuits, as pure inductors and capacitors do not dissipate average power; they only store and release energy.
Important Questions from Alternating Current
In the shown AC source, the voltage is given as V = 20 cos 2000t. Neglecting source resistance, the voltmeter and ammeter readings will be:
A 25 μF capacitor, a 0.10 H inductor, and a 25 Ω resistor is connected in series with an AC source of emf ε = 310 sin 314t. What is the frequency of the AC source?
The same current is flowing in two AC circuits. The first circuit contains a pure inductor and the second, a capacitor. If the frequency of the AC is increased, then the current will:
A 25 μF capacitor, a 0.10 H inductor, and a 25 Ω resistor is connected in series with an AC source of emf ε = 310 sin 314t. What is the frequency of the AC source?
Match List-I with List-II:
List-I List-II (A) Impedance of a series RLC circuit at resonance (I) Voltage across L & C are 180° out of phase (B) For a series LC circuit (II) Current in L & C are 180° out of phase (C) For a parallel LC circuit (III) Minimum (D) Reactance of a capacitor in DC circuit (IV) Infinite Choose the correct answer from the options given below:
