Test: Capacitors & Inductors - 1 - Electrical Engineering (EE) MCQ

Induced voltage across an inductor is zero if the current flowing through it is constant, i.e. Inductor works as a short circuit for DC supply.

Power factor = Real power / Apparent power = kW/kVA

By adding a capacitor in a circuit, an additional kW load can be added to the system without altering the kVA.

Hence, the power factor is improved.

• Q is directly proportional to V.
• The constant of proportionality in this case is C, that is, the capacitance.
• Hence Q=CV. From the given relation we can derive all the equations except for Q = C/V.

Q is directly proportional to V.

The constant of proportionality in this case is C, that is, the capacitance.

Hence Q = CV

From the relation,

C = Q/V = 16/4 = 4F

When capacitors are connected in series, the equivalent capacitance is:
1/C_total=1/C1 + 1/C2

therefore,

C_total = C1*C2 / (C1+C2)

X_L = 2*π*f*L = 10 ohm

Z² = (R²+X_L²)

Therefore the total impedance

Z = 12.2 ohm

V = IZ

therefore,

I= V/Z = 100/12.2 = 8.2A

Due to initial condition, at t = 0 capacitor will act as a constant voltage source (at t = 0, capacitor acts as short-circuit). Hence, option (d) is correct.

∴ Current in the inductor after 3 sec is: 

1= 3 A

Hence, energy is stored after 3 sec:

• Since V is not proportional to R therefore, the element can’t be a resistor.
• At t = 5 ms, even if i ≠ 0, the element behaves as a short circuit therefore, the element can’t be a capacitor (since at t = 0 only capacitor behaves as short circuit).
• The current at t = 0 is zero and at t = 5 ms voltage across the element is zero therefore, the element must be an inductor (at t = 0, an inductor acts as open circuit and at t =∞ it behaves as short circuit).
• From the given voltage and current profile, we have:
  

From given figure:

• The equivalent combination of C₂ and C₃
  
• The equivalent combination of this 1 μF and  C₁ = 2μF is C₁+ 1 μF = 3μF
• Hence, the equivalent capacitance between terminals A and B is
  

Equivalent capacitance between terminals a-b is:

Analysis of AC Circuit with Inductance

• The problem involves an AC circuit where a 220 V alternating current is applied to a pure inductance of 50 H, resulting in a current of 5 A.


• In an AC circuit with only inductance, the voltage leads the current by 90 degrees. This phase difference is crucial to determine the instantaneous values of voltage and current.


 

Calculating Instantaneous Voltage and Current

• The root mean square (RMS) voltage is given as 220 V. The RMS value can be converted to peak voltage (V_m) using the formula: V_m = V_rms × √2.


• Substituting the given values: V_m = 220 V × 1.414 = 311 V (approximately).


 

Understanding Current in Inductive Circuits

• For a pure inductive circuit, the current lags the voltage by 90 degrees. Therefore, if the voltage is described by a sine function, the current will be a sine function shifted by -90 degrees (or equivalently a cosine function).


• The peak current (I_m) is calculated similarly using the RMS current: I_m = I_rms × √2 = 5 A × 1.414 = 7.07 A.


 

Final Formulation

• The instantaneous voltage can be expressed as: v = 311 sin(314t) Volts.


• The instantaneous current, considering the phase shift, is: i = 7.07 sin(314t - 90°) Amps.

 

The phase difference between v and i is: 

Since v leads i therefore, the circuit element is an inductor.

Converting the internal star connected inductance to an equivalent delta, the circuit reduces as shown below.

Hence, equivalent circuit becomes as shown below.