AC Voltage Applied to a Capacitor MCQ Quiz - Objective Question with Answer for AC Voltage Applied to a Capacitor - Download Free PDF

EXPLANATION:

→For circuits consisting of L or C we use the word Reactance in the place of Resistance of the circuit. 

The reactance of a purely inductive circuit is XL = 2πfL

The reactance of a purely capacitive circuit is XC = 1/2πfC

(Where, C = capacitance, L = inductance, f = frequency of the AC circuit)

And a purely resistive circuit does not show any change with changing frequency. 

→Therefore from the above relations we can say,

The reactance of an inductive circuit (X_L) increases with the increasing frequency (f) of the AC circuit. 

∴ the current decreases with increasing frequency (f).

The reactance of a capacitive circuit (X_C) increases with the increasing frequency (f) of the AC circuit. 

∴ the current increases with increasing frequency (f).

∴ The required circuit will be either an RC circuit or a purely capacitive circuit.

So, the correct answers are option (3) and (4).

CONCEPT:

Capacitive reactance:

• The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$

• In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

CALCULATION:

Given C = 100μF = 100 × 10^-6 F, and V = 100 sin(10t)

∵ V = 10.sin(10t)

• The maximum voltage is given as,

⇒ V_max = 100 V     -----(1)

And angular frequency is given as,

⇒ ω = 10 rad/sec     -----(2)

We know that the capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}$     -----(3)

Where ω = angular frequency, and C = capacitance

By equation 2 and equation 3,

$\Rightarrow X_C=\frac{1}{10\times 100\times {10}^{-6}}$

⇒ X_C = 1000 Ω     -----(4)

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$     -----(5)

By equation 1, equation 4, and equation 5, the maximum current in the circuit is given as,

$\Rightarrow I_{max}=\frac{V_{max}}{X_C}$

$\Rightarrow I_{max}=\frac{100}{1000}$

⇒ I_max = 0.1 A

• Hence option 3 is correct.

CONCEPT:

Capacitive reactance:

• The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$

• In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

CALCULATION:

• We know that the capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{2\pi fC}$     -----(1)

Where f = frequency of ac current and C = capacitance

For case 1:

$\Rightarrow X_C=\frac{1}{2\pi fC}$     -----(2)

For case 2: (f' = 2f)

$\Rightarrow X_C^{{}^{\prime }}=\frac{1}{2\pi f^{\prime }C}$

$\Rightarrow X_C^{{}^{\prime }}=\frac{1}{2\pi \times 2fC}$

$\Rightarrow X_C^{{}^{\prime }}=\frac{1}{2}\times \frac{1}{2\pi fC}$     -----(3)

By equation 2 and equation 3,

$\Rightarrow X_C^{{}^{\prime }}=\frac{X_C}{2}$

• Hence option 2 is correct.

CONCEPT:

Capacitive reactance:

• The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$

• In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

CALCULATION:

Given C = 28 μF = 28×10^-6, V = 220 V, and f = 50 Hz

• We know that the capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{2\pi fC}$

$\Rightarrow X_C=\frac{7}{2\times 22\times 50\times 28\times {10}^{-6}}$

$\Rightarrow X_C=\frac{{10}^4}{22\times 4}$

⇒ X_C = 113 Ω

• Hence option 2 is correct.

CONCEPT:

Capacitive reactance:

• The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$

• In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

EXPLANATION:

We know that the capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{2\pi fC}$

$\Rightarrow X_C\propto \frac{1}{f}$     -----(1)

Where f = frequency of ac current and C = capacitance

• By equation 1 it is clear that the capacitive reactance is inversely proportional to the frequency of the AC voltage.
• Therefore if the frequency of the AC voltage is increased in the circuit containing a capacitor, then the capacitive reactance will decrease. Hence option 2 is correct.

Concept:

A circuit that has the only capacitor is known as a purely capacitive circuit.

Explanation:

When the Alternating emf is running in the circuit is

e = e0 sin ωt

Current in the inductive circuit is –

$I=I_0\sin \left(\omega t+\frac{\pi }{2}\right)$

From above it is clear that current leads the voltage by π/2.

The Phase difference between current and voltage in the pure capacitive circuit is 90° or π/2.

A complete cycle makes an angle of 360^o, and the Phase difference between current and voltage is 90^o Hence, current leads the voltage by one-fourth of a cycle

Important Points

When the circuit is purely resistive, inductive, and capacitive.

CONCEPT:

• A circuit that has the only capacitor is known as a purely capacitive circuit.

EXPLANATION:

• Capacitive reactance is defined as the resistance offered to the flow of current by the capacitor.
  • It is denoted by the letter X_L.
• The capacitive reactance is given by:

$\Rightarrow X_c=\frac{1}{\omega C}=\frac{1}{2\pi \nu C}$

Where C = capacitance and f = supply frequency

• Capacitive reactance is inversely proportional to the supply frequency, i.e.

$\Rightarrow X_c\propto \frac{1}{f}$

• If the frequency increases, the capacitive reactance decreases and vice-versa.

• Alternating emf in the circuit is
  $e=e_o\sin \omega t$
• Current in the inductive circuit is –
  $I=I_o\sin \left(\omega t+\frac{\pi }{2}\right)$
• From above it is clear that current leads the voltage by π/2.

CONCEPT:

Capacitive reactance:

• The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$

• In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

EXPLANATION:

We know that the capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{2\pi fC}$

$\Rightarrow X_C\propto \frac{1}{f}$     -----(1)

Where f = frequency of ac current and C = capacitance

• By equation 1 it is clear that the capacitive reactance is inversely proportional to the frequency of the AC voltage.
• Therefore if the frequency of the AC voltage is increased in the circuit containing a capacitor, then the capacitive reactance will decrease. Hence option 2 is correct.

CONCEPT:

• A circuit that has the only capacitor is known as a purely capacitive circuit.

EXPLANATION:

• Capacitive reactance is defined as the resistance offered to the flow of current by the capacitor.
  • It is denoted by letter XC.
• The capacitive reactance is given by:

$\Rightarrow X_c=\frac{1}{\omega C}=\frac{1}{2\pi \nu C}$

Where C = capacitance and ν = supply frequency

• Capacitive reactance is inversely proportional to the supply frequency, i.e.

$\Rightarrow X_c\propto \frac{1}{\nu }$

• If the frequency increases, the capacitive reactance decreases and vice-versa.
• Alternating emf in the circuit is
  $e=e_o\sin \omega t$
• Current in the inductive circuit is –
  $I=I_o\sin \left(\omega t+\frac{\pi }{2}\right)$
• From above it is clear that current leads the voltage by π/2.

CONCEPT:

Energy stored in capacitor:

• A capacitor is a device to store energy.
• The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
• The work done in charging the capacitor is stored as its electrical potential energy.
• The energy stored in the capacitor is

$U=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}C{V}^2=\frac{1}{2}QV$

Where

Q = charge stored on the capacitor,

U = energy stored in the capacitor,

C = capacitance of the capacitor

V = Electric potential difference

Explanation:

From the above explanation, we can see that energy stored in the capacitor is directly proportional to capacitance and square of electric potential.

i.e, $U=\frac{1}{2}C{V}^2$

CONCEPT: ​

• Overall Impedance (Z) in a series combination of an inductor(L) a capacitor (C), and a resistor (R):

$Z=\sqrt{R^2+(\frac{1}{\omega C}-\omega L{)}^2}$

where R is resistance, ωL is the inductance reactance and 1/ωC is Capacitance reactance.

Similarly, the Overall RMS value of voltage is given by:

$E=\sqrt{V_R^2+(V_C-V_L{)}^2}$

where E is the emf, VR is the potential drop across the resistor, VL is the potential drop across the inductance, and VR is the potential drop across the capacitance.

CALCULATION:

Given that VR = 80V; VL = 0; VC = 60V

$E=\sqrt{V_R^2+(V_C-V_L{)}^2}$

$E=\sqrt{{80}^2+(60-0{)}^2}$

E = 100 Volt

So the correct answer is option 4.

CONCEPT:

• A circuit that has the only capacitor is known as a purely capacitive circuit.

• Capacitive reactance (XL) is defined as the resistance offered to the flow of current by the capacitor.

The capacitive reactance is given by:

$\Rightarrow X_c=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where C = capacitance and f = supply frequency

EXPLANATION:​

The alternating emf in the circuit is:
$e=e_o\sin \omega t$

Current in the capacitive circuit is:
$I=I_o\sin \left(\omega t+\frac{\pi }{2}\right)$

From above it is clear that electric current through the capacitor leads the applied voltage by π/2.

So option 1 is correct.

CONCEPT:

• Reactance: It is basically the inertia against the motion of the electrons in an electrical circuit.

• There are two types of reactance:

  1. Capacitive reactance (X_C) (Ohms is the unit)

  2. Inductive reactance (X_L) (Ohms is the unit)

EXPLANATION:

VARIATIONS OF DIFFERENT QUANTITIES WITH FREQUENCY

The variations of reactance with frequency are shown in Fig

Variation of inductive reactance X_L

• The inductive reactance X_L (= 2πfL) is directly proportional to the frequency f, hence its graph is a straight line through the origin.

Variation of capacitive reactance X_C

• The capacitive reactance X_C [= 1/(2πfC)] is inversely proportional to the frequency, hence its graph is a rectangular hyperbola. X_L versus f and X_C versus f curves cut at a point where f = f₀

Variation of net reactance X

• The graph of net reactance x (= X_L - X_C) is obtained from X_L versus f and -X_C versus f curves. It is a hyperbola (but not a rectangular hyperbola). This curve crosses the frequency axis at a point where f = f₀.

CONCEPT:

Capacitive reactance:

• The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$

• In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

CALCULATION:

• We know that the capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{2\pi fC}$     -----(1)

Where f = frequency of ac current and C = capacitance

For case 1:

$\Rightarrow X_C=\frac{1}{2\pi fC}$     -----(2)

For case 2: (f' = 2f)

$\Rightarrow X_C^{{}^{\prime }}=\frac{1}{2\pi f^{\prime }C}$

$\Rightarrow X_C^{{}^{\prime }}=\frac{1}{2\pi \times 2fC}$

$\Rightarrow X_C^{{}^{\prime }}=\frac{1}{2}\times \frac{1}{2\pi fC}$     -----(3)

By equation 2 and equation 3,

$\Rightarrow X_C^{{}^{\prime }}=\frac{X_C}{2}$

• Hence option 2 is correct.

CONCEPT:

Capacitive reactance:

• The capacitive reactance is the opposition offered by the capacitor in an AC circuit to the flow of ac current.
• Its SI unit is Ohm(Ω).
• The capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

Where ω = angular frequency, f = frequency of ac current and C = capacitance

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure capacitor, the capacitive reactance is equal to the impedance.

AC voltage applied to a capacitor:

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$

• In a pure capacitor circuit, the current is ahead of the voltage by one-fourth of a period.

CALCULATION:

Given C = 100μF = 100 × 10^-6 F, and V = 100 sin(10t)

∵ V = 10.sin(10t)

• The maximum voltage is given as,

⇒ V_max = 100 V     -----(1)

And angular frequency is given as,

⇒ ω = 10 rad/sec     -----(2)

We know that the capacitive reactance is given as,

$\Rightarrow X_C=\frac{1}{\omega C}$     -----(3)

Where ω = angular frequency, and C = capacitance

By equation 2 and equation 3,

$\Rightarrow X_C=\frac{1}{10\times 100\times {10}^{-6}}$

⇒ X_C = 1000 Ω     -----(4)

• When an AC voltage is applied to a capacitor, the current in the circuit is given as,

$\Rightarrow I=\frac{V}{X_C}$     -----(5)

By equation 1, equation 4, and equation 5, the maximum current in the circuit is given as,

$\Rightarrow I_{max}=\frac{V_{max}}{X_C}$

$\Rightarrow I_{max}=\frac{100}{1000}$

⇒ I_max = 0.1 A

• Hence option 3 is correct.