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Question

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 correct answer is

50 Hz

Understanding AC Source Frequency from Emf Equation

The question asks for the frequency of the AC source given by the equation \(\varepsilon = 310 \sin 314t\). This equation describes the instantaneous electromotive force (emf) of the source in a series RLC circuit. The standard form for the instantaneous emf of an AC source is:

\(\varepsilon = \varepsilon_0 \sin(\omega t + \phi)\)

Where:

  • \(\varepsilon\) is the instantaneous emf
  • \(\varepsilon_0\) is the peak emf or voltage
  • \(\omega\) is the angular frequency in radians per second (\(\text{rad/s}\))
  • \(t\) is time in seconds
  • \(\phi\) is the phase angle

Comparing the given equation \(\varepsilon = 310 \sin 314t\) with the standard form, we can identify the key parameters:

  • Peak emf, \(\varepsilon_0 = 310 \text{ V}\)
  • Angular frequency, \(\omega = 314 \text{ rad/s}\)
  • Phase angle, \(\phi = 0\)

We are interested in finding the linear frequency of the AC source, which is denoted by \(f\). The relationship between angular frequency (\(\omega\)) and linear frequency (\(f\)) is given by the formula:

\(\omega = 2 \pi f\)

From this relation, we can find the linear frequency \(f\) by rearranging the formula:

\(f = \frac{\omega}{2 \pi}\)

We have the value of \(\omega\) from the given emf equation, which is \(314 \text{ rad/s}\). We also know that the value of \(\pi\) is approximately \(3.14\).

Now, we can substitute the values into the formula to calculate the frequency \(f\):

\(f = \frac{314}{2 \times 3.14}\)

\(f = \frac{314}{6.28}\)

Performing the calculation:

\(f = 50 \text{ Hz}\)

So, the frequency of the AC source is 50 Hz.

Let's quickly look at the options to confirm our result.

  • Option 1: 314 Hz (This is the value of \(\omega\), not \(f\))
  • Option 2: 100 Hz (Incorrect calculation)
  • Option 3: 50 Hz (Matches our calculated value)
  • Option 4: 310 Hz (This is the peak emf \(\varepsilon_0\), not \(f\))

Our calculated frequency of 50 Hz corresponds to Option 3.

Revision Table: Key AC Circuit Concepts

Concept Symbol Relation/Formula Units
Instantaneous Emf (AC Source) \(\varepsilon\) \(\varepsilon = \varepsilon_0 \sin(\omega t + \phi)\) Volts (V)
Peak Emf / Voltage \(\varepsilon_0\) Maximum value of \(\varepsilon\) Volts (V)
Angular Frequency \(\omega\) \(\omega = 2 \pi f\) Radians per second (rad/s)
Linear Frequency \(f\) \(f = \frac{\omega}{2 \pi}\) Hertz (Hz)
Time \(t\) Independent variable in emf equation Seconds (s)

Additional Information on AC Source Frequency

In alternating current (AC) circuits, the voltage and current vary periodically with time. The frequency of the AC source tells us how many complete cycles of variation occur per second.

Angular Frequency (\(\omega\)): This represents the rate of change of the phase angle of the AC wave. It is measured in radians per second. The term \((\omega t + \phi)\) inside the sine function of the emf equation represents the instantaneous phase angle.

Linear Frequency (\(f\)): This represents the number of complete cycles per second. It is measured in Hertz (Hz). 1 Hz means one cycle per second. The standard power line frequency in many parts of the world is 50 Hz or 60 Hz.

The relationship \(\omega = 2 \pi f\) arises because one complete cycle corresponds to a change in phase of \(2 \pi\) radians. If there are \(f\) cycles per second, the total change in phase per second is \(f \times 2 \pi\), which is the angular frequency \(\omega\).

While the given problem provides information about a capacitor, inductor, and resistor, these components are relevant for calculating impedance, current, or phase angle in the circuit. However, the frequency of the AC source is determined solely by the properties of the source itself, as described by its emf equation \(\varepsilon = \varepsilon_0 \sin(\omega t)\).

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Important Questions from Alternating Current

  1. Match List - I with List - II.

    List - IList - II
    (A) The current lags behind the emf by 4π​ in phase(I) A.C. circuit containing only an inductor
    (B) The e.m.f. lags behind the current by 4π​ in phase(II) A.C. circuit with resistance and inductance in series
    (C) The current lags behind the e.m.f. in phase by 2π​(III) A.C. circuit containing only a capacitor
    (D) The e.m.f. lags behind the current in phase by 2π​(IV) A.C. circuit with resistance and capacitor in series

    Choose the correct answer from the options given below:

  2. A e.m.f. es​=50sin314t is applied across a pure capacitor of 637μF. The instantaneous current I is:

  3. Match List - I with List - II.

    List - IList - II
    (A) Resistive Circuit(I) No power dissipation
    (B) Purely inductive or capacitive circuit(II) Maximum power dissipation because XL​=XC​
    (C) LCR series circuit(III) Power dissipated only in resistor
    (D) Power dissipated at resonance in LCR circuit(IV) Maximum power dissipation

    Choose the correct answer from the options given below:

  4. 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:

  5. 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?

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