Alternating Current

Infinite.

The reactance of a capacitor is given by … (i), where ω = angular frequency of oscillation of current, C = capacitance.

For a constant DC source, there are no oscillations in current and hence ω = 0.

Therefore, from (i), reactance X_c becomes = infinite. (Ans)

Zero.

Given:

Voltage

Current

Formula used:

The power dissipated in a series AC circuit is given by

… (i),

where V_rms = root mean square value of voltage,

i_rms = root mean square value of current,

ɸ = phase difference between voltage and current

Now, … (ii), where i₀ = peak value of current, ω = angular frequency of oscillation, t = time

Sincewe can say that the phase difference between the voltage and the current is

Substituting this value in (i), we get

P = 0 (since )

Hence, the power dissipated in the given ac circuit is zero. (Ans)

Equal.

The rms value of current is given by , where = peak value of current.

Since i₁ and i₂ have the same peak value of current = , their rms values will also be equal. (Ans)

Yes.

Let the LCR circuit be connected across an AC supply of voltage … (i)

Now, the impedance in an AC circuit is given by

…. (ii)

where R = resistance, X_L = inductive reactance,

X_c = capacitive reactance,

ω = angular frequency of oscillation of current,

L = inductance,

C = capacitance

Hence, the current in the circuit is given by … (iii),

where V = voltage,

Z = impedance

⇒ … (iv)

Now, at resonance, ⇒ --------------(v)

Substituting (v) in (iv), we get

Current at resonance : … (vi)

Hence, at resonance, voltage across the inductor

⇒ … (vii)

where I_res = current at resonance, X_L = inductive reactance, V = source voltage = V₀sinωt, where V₀ = peak voltage, ω = angular frequency, t = time, R = resistance, L = inductance

Since ω = angular frequency is always greater than equal to 1,

Therefore, voltage across the inductor V_L will be greater than the source voltage V if > 1. (Ans)

No.

Ohm’s law states that the current flowing through a circuit is directly proportional to the potential difference applied across its ends.

Therefore, V = IR, where V = voltage, I = current, R = resistance.

Now, Ohm’s law is valid only in purely resistive circuits (without inductors or capacitors), where there is a linear relationship between voltage and current.

However, Ohm’s law is not valid in case of circuits with non-linear elements, for example those containing inductors or capacitors or a combination of both.

Hence, in the given circuit, Ohm’s law is not consistent. (Ans).

Increase.

The reactance of a capacitor is given by … (i)

where ω = angular frequency of oscillation of current, C = capacitance.

Now, the capacitance of a parallel plate capacitor is given by

(ii), where k = dielectric constant, A = area of plates,

= electric permittivity of vacuum, d = distance between plates.

For vacuum, the dielectric constant k = 1. Let the capacitance in vacuum be … (iii) (from (ii))

For any other medium, k>1. Hence, capacitance of this slab is given by … (iv), where k = dielectric constant, A = area of plates, = electric permittivity of vacuum, d = distance between plates, = capacitance in vacuum

Hence, the reactance of the capacitor in vacuum will be

… (v), where ω = angular frequency, C₀ = capacitance in vacuum

And, the reactance of the capacitor in the dielectric slab will be …(vi) (from (iv)), where k = dielectric constant of slab

Since the dielectric constant k is greater than 1, it becomes clear that X_C1 > X_C2, where X_C1 = reactance of capacitor in vacuum, X_C2 = reactance of capacitor in dielectric slab

Now, the rms value of current is given by

… (vii), where i₀ = peak value of current, V₀ = peak value of voltage, X_C = capacitive reactance

Let the rms value of current initially be i₁ and then be _i2 after insertion of dielectric slab.

Therefore, , where i₀₁ = peak value of current initially, V₀ = peak value of voltage, X_C1 = reactance of capacitor in vacuum

and , where i₀₂ = peak value of current after insertion of slab, V₀ = peak value of voltage, X_C1 = reactance of capacitor after insertion of slab

Since we found out that X_C1> X_C2, it is obvious that

⇒ i₂ > i₁.

Therefore, the rms current increases. (Ans)

No, current will flow through both of them.

At resonance, we know that X_L = X_C⇒ … (i), where X_L = inductive reactance, X_C = capacitive reactance, ω = angular frequency, L = inductance, C = capacitance

Current through an LCR circuit is given by … (ii), where i₀ = peak value of current, V₀ = peak value of voltage, R = resistance, ω = angular frequency, L = inductance, C = capacitance

From (i) and (ii), at resonance, the peak value of current is given by … (iii)

This current will flow through all circuit elements. However, since the inductive reactance and the capacitive reactance are equal, the potential difference across the inductor and capacitor will be equal and opposite and they will cancel each other out. (Ans)

No.

When an AC source is connected to a capacitor, there is a steady-state current in the circuit which transfers charges smoothly between the plates of the capacitor. This results in a potential difference across the plates of the capacitor. The direction of current is alternatively reversed every half-cycle and this is leads to alternating charging and discharging of capacitor.

Same thermal energy, principle of superposition is obeyed.

Given:

Current

Current

Formula used:

The thermal energy produced in one time period due to a current i is given by … (i), where i_rms = rms value of current, R = resistance, ω = angular frequency of oscillation of current

Now, the rms current i_rms in both cases is given by , where i₀ is the peak current.

Therefore, for current i₁, thermal energy produced is

and that produced for current i₂ is also where i₀ = peak value of current, R = resistance, ω = angular frequency of oscillation

Hence, the same thermal energy is produced due to both the currents individually. (Ans)

Since i₁ and i₂ have peak values i₀ and -i₀, they are equal and opposite in value. Hence, the net current through the resistor will be 0 when both pass through the resistor simultaneously. In this case, the thermal energy produced will be 0. (Ans)

Yes, the principle of superposition is obeyed in this case. (Ans)

No.

When a transformer steps up the voltage, the voltage increases but the current decreases in the process since the supplied power remains constant.

Now, the power is given by

where V = voltage, I = current.

Since the voltage and current increase and decrease in proportion to each other, the value of power remains constant.

Hence, energy is not produced. Energy = power x time remains constant. (Ans)

Ideally, we can consider a transformer to be a purely inductive circuit with inductance L.

Hence, the voltage in this case is given by , where i = current, t = time, L = inductance.

⇒

Integrating on both sides, we get

⇒

For a DC source, the current across the inductor increases with time and after a certain amount of time, can reach a very large value. This burns the transformer. (Ans)

(a) No. (b) No.

For an LCR circuit with angular frequency ω, the impedance is given by the formula

where R = resistance, L = inductance, C = capacitance.

Now, the phase difference between the current and the voltage is given by the formula , where = phase difference, ω = angular frequency, L = inductance, C = capacitance, R = resistance.

Since there are no restrictions on the values of L, C or R, it is obvious that can take any value between -∞ to +∞, that is can take any value between -90⁰ to +90⁰ (since tan 90⁰ = ∞)

Since both 120⁰ and 180⁰ fall beyond the permitted range of values, we cannot have an AC circuit with any of the given phase differences. (Ans)

Power after addition of capacitor will decrease. Power after addition of inductor will increase.

The impedance of the circuit after introduction of capacitor is given by … (i), where R = resistance, X_C = capacitive reactance.

Now, average power is given by … (ii), where i_rms = rms value of current, R = resistance

Since i_rms decreases with increase in impedance, hence on the introduction of a capacitor, the average power absorbed by the resistance will also decrease. (Ans)

Now, the impedance of an LCR circuit is given by … (iii), where R = resistance, X_L = inductive reactance, X_C = capacitive reactance.

Hence, compared to Z₁, the value of Z₂ is less and the rms value of current is greater in this case. Therefore, if a small inductance is also introduced in the circuit, the average power absorbed increases. (Ans)

(i) Yes (ii) No

A hot wire ammeter only measures the root mean square(rms) value of alternating current. Hence, when it is used to measure a direct current, it will not show the fluctuating value of ac, but only show a constant current equal to the rms value of the current. Thus, it can be used to measure direct current having constant value. (Ans)

No, we do not need to change the graduations since the rms value of current is the same as the direct current.

The resistance of a capacitor is given by , where ω = angular frequency of oscillation of current C = capacitance.

Now, in case of DC current, ω = 0.

Hence, the resistance for DC becomes = infinite. (Ans)

Given:

Emf

Steady state current

Formula used:

Charge in steady state will be given by

… (i),

where C = capacitance, ε = emf, t = time

Hence, current is given by … (ii), where Q = charge, t = time

⇒ , from (i)

Comparing this with _, we get and .

Hence, we find that i₁< i₂ (Ans).

Given:

Rms value of AC source voltage V_rms = 220 V

Hence, peak value of voltage = V which is around 310V. (Ans)

The frequency of the AC source is 50 Hz.

which means the time period of the wave is, 1/50 = 002 sec

Now the average voltage of the wave at the time interval 0.01 sec

If the interval lies between π/2 and 3π/2, the average voltage will be zero. In all the other case it will not be zero.

Hence, it may be zero. Thus B is the correct answer.

The magnetic field energy in an inductor is given by … (i), where L = inductor, i = current.

The graph of alternating current is given by:

From (i), we can see that the magnetic field energy reached its maximum and minimum value when the current is maximum and 0 respectively.

Also, from the graph we can see that the time taken by the current to change from its maximum to zero value is T/4, where T = time period.

Now, from the given problem, = 5 ms = 0.005 s

Hence, time period T = (0.005 x 4) s = 0.02 s

Therefore, frequency of oscillation is equal to = = 50 Hz. (Ans)

The reactance of a series LC combination is given by

where X_L = resistance of inductor, X_C = resistance of capacitor, f = frequency, L = inductance, C = capacitance.

Also, we can see that when X = 0(point intersecting on the graph), .

This is correctly represented in graph (d). (Ans)

The impedance of an AC circuit is given by … (i), where R = resistance, X = impedance.

Given: Resistance R = 4 Ω and Impedance X = 3 Ω

Substituting these values in (i), we get

Impedance Z = = 5 Ω (Ans)

Transformers only work in AC circuits where they step up or step down the voltage.

In a DC circuit, there is no change in flux with time across the coils of the conductor, since the current is constant. So, there will be no induced emf in the secondary coil due to the change in flux from changing current in primary coil. For this reason, DC circuits do not obey the principle of transformers.

The rms value of current is given by

… (i), where

i = current = i₁ cos ωt + i₂ sin ωt …. (ii)(given), t = time, ω = angular frequency, T = time period.

Now, squaring on both sides of (ii), we get

i² = i₁²cos²ωt + 2i₁i₂sinωtcosωt + i₂²sin²ωt … (iii)

=

ow, … (v), where ω = angular frequency, T = time period = 2π/ω

Therefore, (iv) becomes:

=> [since sin nπ = 0, cos 0 = cos 2nπ = 1]

Hence, rms value of current i is (Ans)

To produce the same heating effect, the constant current required(i) will be the root mean square(rms) current(irms)

Hence, where i₀ = peak current = 14 A

=>i = (14/√2) A = 9.899 A which is about 10 A. (Ans)

The rms current is equal to the value of the constant current.

Hence, the rms current is also 2.8 A. (Ans)

The reactance of an inductor is given by , where f = frequency, L = inductance.

Hence, if the frequency is increased, the reactance of the inductor also increases. Therefore, option (A) is correct. (Ans)

The resistance remains unchanged with change in frequency. Hence, option (B) is incorrect.

The reactance of a capacitor is given by , where f = frequency, C = capacitance.

Hence, if the frequency increases, the reactance of the capacitor decreases. Hence, option (C) is incorrect.

Now, the reactance of the circuit is given by

Since on increasing the frequency, X_L increases but X_C decreases, their overall difference, that is reactance of the circuit, also increases. Hence option (D) is also correct. (Ans)

The reactance of a circuit is given by , where X_L = reactance of inductor, X_C = reactance of capacitor.

X can be 0 in two cases:

(i) When , that is both inductor and capacitor are present. Hence option (A) is correct. (Ans)

(ii) When both X_L and X_C = 0, that is, there is neither an inductor nor a capacitor. Hence, option (D) is correct. (Ans)

When either an inductor or a capacitor is present, either X_L or X_C are non-zero and hence the reactance cannot be 0. Therefore, options (B) and (C) are incorrect.

In a pure inductive circuit, the voltage leads the current by a phase difference of 90⁰. Hence, when the instantaneous voltage is maximum, the current is zero and vice versa. Hence, option (A) is correct. (Ans)

In a pure capacitive circuit, the voltage lags behind the current by a phase difference of 90⁰. Hence, when the instantaneous voltage is maximum, the current is zero and vice versa. Hence, option (B) is correct. (Ans)

In case of a combination of an inductor and a capacitor also, the current may lead or lag behind the voltage by 90⁰, depending on whether the voltage across the inductor or capacitor is greater. Hence, when the instantaneous voltage is maximum, the current is zero and vice versa. Hence, option (D) is correct. (Ans)

Option (C) is incorrect because in a pure resistor circuit, the current and voltage are in phase with each other. Hence, when the voltage is maximum, the current is also maximum and vice versa.

For an inductor-coil circuit with some resistance, the current and the induced emf in the inductor are of the sinusoidal form. It is shown in the diagram below:

Here, T is the time period.

From the graph, we can see that the average value of current or induced emf over a cycle is 0.

Mathematically, we can see it in the following way:

Let the emf be of the form E = E₀sinωt, where E₀ = peak value of emf, ω = angular frequency, t = time.

Average emf over an entire cycle

Where T = time period

Similarly, we can also show that the average value of current over a full time period is also 0.

Hence, options (A) and (B) are correct.

Joule heat is given by H = i_rms² x R, where i_rms = rms value of current, and R = resistance, which is non zero. Hence, option (C) is incorrect.

Magnetic energy stored in inductor is given by , where L = inductance, i_rms = rms value of current, which is non zero. Hence, option (D) is incorrect.

Only a hot-wire voltmeter can be used to measure AC voltage across a resistance. Normal ammeters cannot be used for this purpose due to the changing value and direction of alternating current. All other devices can measure only the DC voltage. Hence only option (B) is correct. (Ans)

Both DC and AC dynamo can be used to convert mechanical energy into electrical energy using wire coils rotating in a magnetic field. Hence, options (A) and (B) are correct. (Ans)

A motor is used to convert electrical energy to mechanical energy and not the other way around. Hence, option (C) is incorrect.

A transformer is used to step up or step down the voltage or to transfer electrical energy only. It cannot be used for transformation of energy from one type to another. Hence, option (D) is also incorrect.

Given:

Rms value of voltage(V_rms) = 100 V

Rms value of current(i_rms) = 10 A

Formula used:

The average power is given by, where V_rms = rms value of voltage, i_rms = rms value of current, = phase difference between current and voltage.

Substituting the given values, we get

Now, cos can have any value between 0 to 1. (since can have any value between )

Therefore, the range of values for P is:

0 ≤ P ≤ 1000 W.

Therefore, the power can be less than 1000 W, or may be equal to 1000 W. Hence, options (B) and (D) are correct. (Ans)

Options (A) and (C) are incorrect since P can have any value between 0 and 1000 W, and can never be greater than 1000 W.

I is current at any time ‘t’,

I₀ is maximum value of the current in the circuit,

F is the frequency of the alternating current=

Current at any time is given as

Given that

Frequency= 50 Hz

Then, Peak voltage ( is given as

Now, time taken for the current to reach the peak value= time taken to reach the zero value from rms

Where, f is the frequency, t is time taken and is angular velocity.

Given that: Power (P) = 60W

Alternate Voltage (V) =220V

Now, power can be expressed as

Therefore,

Then instantaneous voltage is

Thus, the maximum instantaneous current through the filament

Given that an electric bulb is designed to operate at voltage = 12V.

If the bulb is connected to an AC source and given normal brightness, then the peak voltage will be

Given that: Peak power ( = 80 W

Then, instantaneous power is

The energy consumed by the coil in time t=100 seconds will be

= P × t = 40 × 100

= 4000J

Energy consumed = 4.0KJ

Given: Dielectric strength of air (E) = 3.0 × 10⁶ V/m,

Area (A) =20 cm² and separation width (d) = 0.10 mm.

Potential difference (V) across the capacitor is

The maximum rms voltage of an AC source which can be safely connected to this capacitor will be given as

The current in a discharging LR circuit is given by

Then, the rms current for the period t=0 to t= T can be obtained by:

So the rms current is

Capacitance of the capacitor C=10 μF,

Output voltage of the oscillator ϵ = (10V) sin ωt.

On comparing the output voltage of the oscillator with

We get Peak voltage

For a capacitive circuit,

Reactance,

Here, is angular frequency,

C is capacitance of capacitor,

Peak current

{a} At

Peak current

{b} At

Peak current

{c} At

Peak current

{d} At

Peak current

Given: Inductance of a coil= 5.0 mH and

Peak voltage

{a} At ω= 100 s^–1

Reactance of a coil is given by

Then,

Here is angular velocity

Peak current

{b} At ω= 500 s^–1

Reactance of a coil is given by

Then,

Here is angular velocity

Peak current

{c} At ω= 1000 s^–1

Reactance of a coil is given by

Then,

Here is angular velocity

Peak current

Given: Resistance (R) = 10 Ω,

Inductance (L) = 0.4 Henry,

It is connected to an AC source (E) of 6.5 V and

Frequency (f) = 30/π Hz.

Then, Impedance (Z) of a coil is given by

Hence, the rms current will be

And

Thus, the average power consumed in the circuit will be

W

Given: Resistance (R) =100 Ω

Source emf ϵ = (12 V) sin (250 π s^–1) t

T=1.0 ms =10^-3 s.t

Then, energy dissipated (E) will be

Since,

On integrating,

Given: Resistance (R) = 300Ω,

Capacitance (C) = 25 μF =,

Rms voltage (ϵ₀) = 50 V and

Frequency (v) = 50/π Hz.

Then, Reactance will be

Now, Impedance (Z) will be

So, Peak current will be

And average power dissipated will be

Given: Power (P) = 55W,

Bulb operated at voltage (V) = 110V,

Voltage supplied (E) = 220V

Resistance (R) will be

Frequency (f) =50Hz,

Angular velocity will be

Current (I) in the circuit will be

Voltage drop across the resistor will be

Where, L is the inductance of the coil for which the bulb gets correct voltage.

Given: resistance R = 300Ω,

Capacitance of capacitor C = 20 μF,

Inductance of inductor L = 1.0 Henry,

Voltage across the circuit ϵ_rms = 50 V and

Frequency v = 50/π Hz

{a} rms current in the circuit will be

Where Z is impedance in the circuit

Then,

{b} potential difference across the capacitor will be

Potential difference across the resistor will be

Potential difference across the inductor will be

Net sum of all potential drops =

Rms voltage,

Hence, sum of all potential drops > rms potential applied.

Given: resistance R = 300Ω,

Capacitance of capacitor C = 20 μF,

Inductance of inductor L = 1.0 Henry,

Voltage across the circuit ϵ_rms = 50 V and

Frequency v = 50/π Hz

Then the rms current across the circuit

Where Z is impedance in the circuit

Then,

Electric energy stored in capacitor will be

Magnetic field energy stored in the coil will be

{a} for current to be maximum in a circuit

Where, f is frequency, is angular velocity, L is inductance of inductor, C is capacitance of capacitor, is resonance across inductor and is resonance across capacitor.

{b} maximum current (I) will be

A

Given that rms voltage =24V

Internal resistance, r= 4 ohm

Rms current =6A

Then, rms resistance will be

If this inductor coil is connected to a battery of emf E= 12 V and internal resistance r’ = 4.0Ω, then the steady current will be

Since, net resistance R’ = (R + r’) = 4+4 =8 ohm.

Therefore, the steady current (I) is

Given: Voltage V₁=10×10^-3V,

Resistance (R) =1×10³ ohm,

Capacitance (C) =10×10^-9 F

Angular velocity will be

{a} At frequency (f) =10kHz

Reactance,

Impedance,

Then, Current (I₀) will be

Thus, Output voltage (V₀) will be

{b} At frequency (f) =100kHz

Reactance,

Impedance,

Then, Current (I₀) will be

Thus, Output voltage (V₀) will be

{c} At frequency (f) =1 MHz

Reactance,

Impedance,

Then, Current (I₀) will be

Thus, Output voltage (V₀) will be

{a} At frequency (f) =10 MHz

Reactance,

Impedance,

Then, Current (I₀) will be

Thus, Output voltage (V₀) will be

Given that a transformer has 50 turns in the primary and 100 in the secondary.

If the primary is connected to a 220 V DC supply, the voltage across the secondary is zero because a transformer does not work on DC.

The transformer works on the principle of mutual induction, for which current in one coil must change uniformly. If DC supply is given, the current will not change due to constant supply and the transformer will not work.