Alternating Current — Page 5 | JEE Physics

Q41

A parallel plate capacitor has a capacitance

\mathrm{C}=200~ \mathrm{pF}

. It is connected to

230 \mathrm{~V}

ac supply with an angular frequency

300~ \mathrm{rad} / \mathrm{s}

. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

A

14.3 ~\mu \mathrm{A}

and

143 ~\mu \mathrm{A}

B

13.8 ~\mu \mathrm{A}

and

13.8 ~\mu \mathrm{A}

C

13.8 ~\mu \mathrm{A}

and

138 ~\mu \mathrm{A}

D

1.38 ~\mu \mathrm{A}

and

1.38 ~\mu \mathrm{A}

Correct Answer

Option B

Solution

To solve this problem, we need to understand how to calculate the rms current in an AC circuit containing a capacitor, as well as the concept of displacement current.

First, the rms (root mean square) value of the current ( $I_{\text{rms}}$ ) in a capacitor when connected to an AC supply is given by: $I_{\text{rms}} = V_{\text{rms}} \cdot \omega C$ where $V_{\text{rms}}$ is the rms voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.

For an AC supply, the rms voltage is related to the peak voltage ( $V_0$ ) by: $V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}}$ However, we've been given the rms voltage directly, which is $230\, \text{V}$ .

Therefore, we can use this value directly in our calculations.

Given: $C = 200\, \text{pF} = 200 \times 10^{-12}\, \text{F}$ $\omega = 300\, \text{rad/s}$ $V_{\text{rms}} = 230\, \text{V}$ Substitute these values into the equation for rms current: $I_{\text{rms}} = V_{\text{rms}} \cdot \omega C = 230 \times 300 \times 200 \times 10^{-12} = 13.8 \times 10^{-6}\, \text{A}$ Therefore, the rms value of the conduction current is $13.8\, \mu \text{A}$ .

Now, for the displacement current (which is essentially the current that would "flow" through the dielectric of the capacitor as a result of the changing electric field).

In an AC circuit, the displacement current and the conduction current are the same.

That's because the displacement current is necessary to sustain the changing electric field between the plates of the capacitor, and this changing field is what causes the conduction current in the leads and the rest of the circuit.

Hence, the rms value of the displacement current is the same as the conduction current, which is $13.8\, \mu \text{A}$ .

So, the correct option is: Option B $13.8\, \mu \text{A}$ and $13.8\, \mu \text{A}$

Q42

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb :

A becomes zero

B remains same

C increases

D decreases

Correct Answer

Option C

Solution

To understand the impact of placing a dielectric between the plates of the capacitor on the glow of the bulb, we need to consider the properties and behavior of capacitors in an AC circuit.

When a capacitor is connected in series with a bulb in an AC circuit, the impedance of the capacitor plays a significant role in determining the current through the circuit.

The impedance $Z_C$ of a capacitor in an AC circuit is given by:

Z_C = \frac{1}{\omega C}

where: $Z_C$ is the capacitive reactance (impedance of the capacitor). $\omega$ is the angular frequency of the AC supply ( $\omega = 2 \pi f$ , where $f$ is the frequency). $C$ is the capacitance of the capacitor.

When we place a dielectric between the plates of the capacitor, the capacitance $C$ increases.

The capacitance with a dielectric can be described as:

C' = \kappa C

where: $C'$ is the new capacitance with the dielectric present. $\kappa$ is the dielectric constant ( $\kappa > 1$ ).

Since $C' > C$ , the new capacitive reactance $Z_C'$ can be given as:

Z_C' = \frac{1}{\omega C'}

Because $C' = \kappa C$ , we have:

Z_C' = \frac{1}{\omega \kappa C} = \frac{1}{\kappa} \left( \frac{1}{\omega C} \right) = \frac{Z_C}{\kappa}

Since $\kappa > 1$ , $ Z_C' Thus, with an increase in current, the bulb will glow brighter.

Therefore, the correct option is: Option C: increases

Q43

In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to : A. pure inductor. B. pure capacitor. C. pure resistor. D. combination of an inductor and capacitor. Choose the correct answer from the options given below :

A A, B and C only

B A and B only

C A, B and D only

D B, C and D only

Correct Answer

Option C

Solution

To understand which answer is correct, we must consider the relationship between voltage and current in various types of circuits, namely circuits with pure inductors, pure capacitors, pure resistors, and a combination of inductors and capacitors (LC circuits).

In a purely resistive circuit, the voltage and current are in phase, meaning when the voltage is maximum, the current is also maximum.

Therefore, option C (pure resistor) cannot result in the instantaneous current being zero when the instantaneous voltage is maximum.

In a purely inductive circuit, the current lags the voltage by $90^\circ$ , or in other words, when the voltage is at its maximum, the current is zero.

This is because the inductor opposes changes in current, leading to this phase difference.

In a purely capacitive circuit, the current leads the voltage by $90^\circ$ .

This means when the voltage is at its maximum value, the current through the capacitor is zero because the current reaches its maximum or minimum before the voltage does.

In an LC circuit (inductor-capacitor), under certain conditions like at its resonant frequency, the circuit behaves as if it is purely resistive in nature where voltage and current are in phase.

However, it's more nuanced because the question likely implies a condition not at resonance but rather at a general characteristic of LC circuits.

In LC circuits, aside from the resonance condition, the presence of both inductive and capacitive components can lead to scenarios where the inductive and capacitive reactances cancel each other, particularly in the case where oscillations are involved, and indeed, there can be moments when the voltage is maximum and the current is minimum (zero in the ideal case) due to the energy being alternately stored in the magnetic field of the inductor and the electric field of the capacitor.

Thus, reflecting on the scenarios: Pure inductor - Fits the description: instantaneous current is zero when the instantaneous voltage is maximum.

Pure capacitor - Also fits the description for the reason described, taking into account ideal conditions.

Pure resistor - Does not fit as voltage and current are in phase.

Combination of an inductor and capacitor (LC circuit) - Can fit the description under certain non-resonance conditions where the behavior of energy exchange between L and C at a specific instant can result in the instant current being zero when the voltage is maximum, but this is more complex and dependent on specific conditions unlike the straightforward lag or lead in purely inductive or capacitive circuits.

Therefore, the correct answer is Option C: A, B, and D only.

Q44

A coil of negligible resistance is connected in series with

90 \Omega

resistor across

120 \mathrm{~V}, 60 \mathrm{~Hz}

supply. A voltmeter reads

36 \mathrm{~V}

across resistance. Inductance of the coil is :

A 0.91 H

B 0.76 H

C 2.86 H

D 0.286 H

Correct Answer

Option B

Solution

To find the inductance of the coil, we need to analyze the given circuit and use the information provided.

The circuit consists of a resistor and an inductor in series, connected to an AC supply.

Here are the given values: 1.

Resistance,

R = 90 \Omega

2. Supply voltage,

V_{\text{total}} = 120 \mathrm{~V}

3. Frequency,

f = 60 \mathrm{~Hz}

4. Voltage across the resistor,

V_R = 36 \mathrm{~V}

First, we calculate the current through the resistor (which is the same as the current through the inductor, since they are in series) using Ohm's law:

I = \frac{V_R}{R} = \frac{36}{90} = 0.4 \mathrm{~A}

Next, we find the total impedance

Z

of the series combination from the total supply voltage:

V_{\text{total}} = I \cdot Z

Z = \frac{V_{\text{total}}}{I} = \frac{120}{0.4} = 300 \Omega

We know that the total impedance in a series circuit consisting of a resistor and an inductor is given by:

Z = \sqrt{R^2 + (X_L)^2}

where

X_L

is the inductive reactance. Rearrange this to solve for

X_L

:

X_L = \sqrt{Z^2 - R^2} = \sqrt{(300)^2 - (90)^2} = \sqrt{90000 - 8100} = \sqrt{81900} \approx 286 \Omega

Now, we use the inductive reactance formula to find the inductance

L

:

X_L = 2\pi f L

L = \frac{X_L}{2\pi f} = \frac{286}{2 \pi \cdot 60} \approx \frac{286}{376.99} \approx 0.76 \mathrm{~H}

Thus, the inductance of the coil is: Option B: 0.76 H

Q45

An alternating current is represented by the equation,

i=100 \sqrt{2} \sin (100 \pi t)

ampere. The RMS value of current and the frequency of the given alternating current are

A

\dfrac{100}{\sqrt{2}} \mathrm{~A}, 100 \mathrm{~Hz}

B

50 \sqrt{2} \mathrm{~A}, 50 \mathrm{~Hz}

C

100 \sqrt{2} \mathrm{~A}, 100 \mathrm{~Hz}

D

100 \mathrm{~A}, 50 \mathrm{~Hz}

Correct Answer

Option D

Solution

To solve the problem, we need to determine both the RMS value of the current and the frequency from the given equation:

i(t)=100\sqrt{2}\sin(100\pi t).

Here's how we do it step by step: RMS Current Calculation: For a sinusoidal current of the form

i(t)=I_0 \sin(\omega t),

the RMS (root-mean-square) current is given by:

I_{\text{RMS}} = \frac{I_0}{\sqrt{2}}.

In our case, the amplitude

I_0 = 100\sqrt{2}\, \text{A}.

Plugging into the formula:

I_{\text{RMS}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\, \text{A}.

Frequency Calculation: The argument of the sine function is

100\pi t.

In the general form

\sin(\omega t),

$\omega$ is the angular frequency which relates to the frequency

f

by the equation:

\omega = 2\pi f.

Given

\omega = 100\pi,

we can solve for

f

:

f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50\, \text{Hz}.

With these calculations, we find that the RMS current is

100\, \text{A}

and the frequency is

50\, \text{Hz}.

Thus, the correct option is: Option D:

100\,\text{A}, \, 50\,\text{Hz}.

Q46

An electric bulb rated as

100 \mathrm{~W}-220 \mathrm{~V}

is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :

A 0.32 A

B 0.64 A

C 0.45 A

D 2.2 A

Correct Answer

Option B

Solution

To find the peak current through the bulb, follow these steps: Step 1: Calculate the bulb's rms (root mean square) current: The formula to find the rms current is:

I_{\text{rms}}=\frac{P}{V}

Here, $P = 100 \text{ W}$ and $V = 220 \text{ V}$ . Substitute these values into the formula:

I_{\text{rms}} = \frac{100\text{ W}}{220\text{ V}} \approx 0.455\text{ A}

Step 2: Convert the rms current to peak current: Use the following formula to find the peak current:

I_{\text{peak}} = I_{\text{rms}} \times \sqrt{2}

Substitute the value of $I_{\text{rms}}$ into the formula:

I_{\text{peak}} = 0.455 \times 1.414 \approx 0.64\text{ A}

The correct choice is 0.64 A (Option B).

Q47

Match List - I with List - II : .tg .tg List I List II A. AC generator I. Presence of both L and C B. Transformer II. Electromagnetic Induction C. Resonance phenomenon to occur III. Quality factor D. Sharpness of resonance IV. Mutual Induction Choose the correct answer from the options given below :

A A-II, B-I, C-III, D-IV

B A-IV, B-III, C-I, D-II

C A-II, B-IV, C-I, D-III

D A-IV, B-II, C-I, D-III

Correct Answer

Option C

Solution

AC generator works on EMZ principle (A-II) Transformer uses Mutual induction (B-IV) Resonance occurs when both $L$ and $C$ are present (C-Z) and quality factor determines sharpness of resonance (D-III)

Q48

Match with . .tg .tg

List - I		List - II
(A)	AC generator	(I)	Detects the presence of current in the circuit
(B)	Galvanometer	(II)	Converts mechanical energy into electrical energy
(C)	Transformer	(III)	Works on the principle of resonance in AC circuit
(D)	Metal detector	(IV)	Changes an alternating voltage for smaller or greater value

A (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

B (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

C (A) - (III), (B) - (IV), (C) - (II), (D) - (I)

D (A) - (III), (B) - (I), (C) - (II), (D) - (IV)

Correct Answer

Option A

Solution

.tg .tg AC generator $\to$ Converts mechanical energy into electrical energy Galvanometer $\to$ Detects the presence of current in the circuit Transformer $\to$ Change AC voltage for smaller or greater value Metal detector $\to$ Works on the principle of resonance in AC circuit

Q49

A series AC circuit containing an inductor (20 mH), a capacitor (120

\mu

F) and a resistor (60

\Omega

) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is :

A 5.65

\times

102J

B 2.26

\times

103J

C 5.17

\times

102 J

D 3.39

\times

103 J

Correct Answer

Option C

Solution

Energy dissipated in 60 Sec = (Pavg) $\times$ 60 = Vrms $\times$ Irms $\times$ cos $\phi$ $\times$ 60 = Vrms $\times$

{{{V_{rms}}} \over Z} \times

cos $\phi$ $\times$ 60 XL = $\omega$ L = 2 $\pi$ FL = 2 $\pi$ (50) $\times$ 20 $\times$ 10 $-$ 3 = 2 $\pi$

\Omega

XC =

{1 \over {\omega C}}

=

{1 \over {2\pi fC}}

=

{1 \over {2\pi \left( {50} \right) \times 120 \times {{10}^{ - 6}}}}

= 26.52

\Omega

$\therefore$ XC $-$ XL = 20.24

\simeq

20 $\therefore$ Z =

\sqrt {{{\left( {{X_C} - {X_L}} \right)}^2} + {R^2}}

=

\sqrt {{{\left( {20} \right)}^2} + {{60}^2}}

= 20

\sqrt {10}

\Omega

Also cos $\phi$ =

{R \over Z}

=

{{60} \over {20\sqrt {10} }} = {3 \over {\sqrt {10} }}

$\therefore$ Energy dissipated in 60 sec =

{{{{\left( {24} \right)}^2}} \over {20\sqrt {10} }} \times {3 \over {\sqrt {10} }} \times 60

= 5.17 $\times$ 102 J

Q50

A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :

A 50 A

B 45 A

C 35 A

D 25 A

Correct Answer

Option B

Solution

Given, Primary voltage (VP) = 2300 V Primary current (IP) = 5A Secondary voltage (VS) = 230 V efficiency ( $\eta$ ) = 90% We know, Efficiency ( $\eta$ ) =

{{{\mathop{\rm Sec}\nolimits} ondary\,Power} \over {\Pr imary\,\,Power}}

$\therefore$ $\eta$ =

{{{P_S}} \over {{P_P}}}

=

{{{V_S}\,{I_S}} \over {{V_P}\,{I_P}}}

$\Rightarrow$ 0.9 =

{{230 \times {{\rm I}_S}} \over {2300 \times 5}}

$\Rightarrow$ I = 45 A