Test: Capacitors & Inductors - 1

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:

Max. value of current:

Assuming voltage as the reference phasor:

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.