Physics M.C.Q [1M]

3. R = 15Ω, L = 3.5 H, C = 30μF.

Solution:

Quality factor (Q) of an L-C-R circuit is given by, $\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}$

Tuning of an L-C-R circuit depends on quality factor of the circuit. Tuning will be better when quality factor of the circuit is high.

As, quality factor (Q) of an L-C-R circuit is given by, $\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}$

For Q to be high, R should be low, L should be high and C should be low, Therefore option (c) is most apporopriate.

1. $5\Omega$

Explanation:

Impedance(Z) $=\sqrt{\text{R}^2+\text{X}^2}$

where, R = resistance

X = reactance

Given $\text{R}=4\Omega\text{ and }\text{X}=3\Omega$

Substitute values back in equation

$\text{Z}=\sqrt{4^2+3^2}$

$\text{Z}=5\Omega$

3. 1.2 × 10^-4

Explanation:

In a series LCR circuit, resonance occurring at 105Hz. At that time, the potential difference across the 100 resistance is 40V while the potential difference

across the pure inductor is 30V. The inductance L of the inductor is equal to

Current(i) $=\frac{\text{V}_\text{r}}{\text{R}}=\frac{100}{40}=2.5\text{A}$

the inductor value is V_L ​= i × (wL)

$\text{L}=\frac{30}{2.5\times2\times\pi1\times105}=1.2\times10^{-4}\text{H}$

1. For a given power level, there is a lower current.
2. Lower current implies less power loss.

4. It is easy to reduce the voltage at the receiving end using step-down transformers.

Solution:

Key Concept: Power loss due to transmisssion lines having resistance (R) and I_rms current flowing in the circuit is I²_rms R.

The power is to be transmitted over the large distances ar high alternating voltages, so current flowing through the wires will be low because of given power (P).

For a given power level, we find that

P = E_rmsI_rms (I_rms is low when E_rms is high)

Powor loss = I²_rms R = low ($\because$ I_rms is low)

Now at the receiving end high voltage is reduced by using step-down transformers.

2. $\text{ohm}^{-1}$

Explanation:

Susceptance is the imaginary part of admittance. 

The admittance is the inverse of impedance. 

Unit of admittance is ohm. Unit of admittance is ohm^-1 or siemens. 

Unit of susceptance is same as of admittance.

1. $\frac{10}{2\pi}\text{Hz}$

Explanation:

Resonant frequency $=\frac{1}{2\pi\sqrt{\text{LC}}}$

$=\frac{1}{2\pi\sqrt{10^{-6}}}$

$=\frac{10}{2\pi}\text{Hz}$

4. none of the above is true

Explanation:

The voltages across different elements in AC circuit add vectorially.

The voltages across inductor and capacitor are out of phase and at 90^∘ to that across resistor.

Hence net voltage across source $=\sqrt{(\text{V}_1-\text{V}_2)+\text{V}^2_3}.$

4. Only L

Explanation:

In an inductor, current lags behind the input voltage by a phase difference of $\frac{\pi}{2}$.

Current and voltage are in same phase in resistor whereas current leads the voltage by $\frac{\pi}{2}$in a capacitor.

So, the circuit must contain an inductor only.

2. May be 1000W.

4. May be less than 1000W.

Explanation:

Average power $\text{P}_\text{av}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\phi$

$=100\times10\cos\phi$

$\text{P}_\text{av}=1000\cos\phi$

$\therefore\ \cos\phi$ lies "0 to 1".

$\Rightarrow\ 0\leq\text{P}_\text{av}\leq1000.$

1. An inductor and a capacitor.

Explanation:

$\text{X}=0$ (Given)

$\text{X}=\text{X}_\text{L}+\text{X}_\text{C}$

$=\omega\text{L}-\frac{1}{\omega\text{C}}=0$

It is possible that the circuit contains an inductor and a capacitor.

1. Of the inductor increases.

Explanation:

$\text{X}_\text{L}=\omega\text{L}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}$

If frequency increases that causes 'X_L' raction of inductor increases and 'X_C' reactance of capacitor decreses.

4. About 10A.

Explanation:

$\text{I}_\text{p}=14\text{Amp}$

$\text{I}_\text{rms}=\frac{\text{IP}}{\sqrt{2}}=\frac{14}{\sqrt{2}}$

$=9.9=10\text{Amp}.$

2. $10\sqrt{\text{2}}\text{mA}$

Explanation:

$\text{i}=\frac{\text{V}_\text{rms}}{\sqrt{\text{R}^2+\Big(\frac{1}{\omega\text{C}}-\omega\text{L}\Big)^2}}$

$\Rightarrow\text{i}=\frac{200\sqrt{2}\frac{1}{\sqrt{2}}}{\sqrt{10000^2+\Big(\frac{1}{100\times10^{-6}}-100\times0\Big)^2}}=\frac{200}{\sqrt{2\times10000^2}}$

$10\sqrt{\text{2}}\text{mA}$

3. $13.2\Omega$

1. 2

Explanation:

$\text{I}=\frac{\text{E}}{\sqrt{\text{R}^2+(\text{X}_\text{L}-\text{X}_\text{C})^2}}$

$=\frac{110}{\sqrt{55^2+(2-2)^2}}$

$=\frac{110}{55}=2\text{A}$

3. 225.08Hz

Explanation:

The resonant frequecy of a circuit is given by $\text{f}=\frac{1}{2\pi\sqrt{\text{LC}}}$

Given $\text{L}=0.1\text{H }\text{C}=5\mu\text{F }\text{R}=5\Omega$

Substituting them in formula for f gives

f = 225.08 Hz.

2. $157\Omega$

Explanation:

$\text{Reactance }=\omega\text{L}$

$=2\pi\text{fL}$

$=2\pi\times50\times0.5$

$157\Omega$

3. Admittance

Explanation:

Impedance is the opposition a circuit presents to a current when a voltage is applied.

Admittance is a measure of how easily a circuit or device will allow a current to flow.

Admittance is defined as $\text{Y}=\frac{1}{\text{Z}}$

where Z is the impedance of the circuit.

4. lags behind the voltage by $\frac{\pi}{2}$

Explanation:

In a purely inductive circuit (an AC circuit containing inductance only) the current lags behind the voltage by $\frac{\pi}{2}$

3. 1.0012

Explantion:

We know frequency is given by:

$\text{n}_1=\frac{1}{2\pi\sqrt{\text{LC}_1}}$

And

$\text{n}_2=\frac{1}{2\pi\sqrt{\text{LKC}_1}}$

Thus we get the ratio of frequencies as

$\frac{\text{n}_1}{\text{n}_2}=\frac{150000}{149900}=\sqrt{\text{k}}$

Solving the above equation we get, $\text{K}\approx1.0012$

1. $12\Omega$

Explanation:

I_rms​ = currentincircuit = 3A

$\text{p}=108\text{W}=\text{I}_\text{ems}^2\text{R}=3^2\text{R}$

$\text{R}=\frac{108}{9}=12\Omega$

3. $\frac{\text{E}_\text{v}}{\sqrt{\text{R}^2+\omega^2\text{L}^2}}$

Explanation:

The impedance in R-L circuit is 

$\text{Z}=\sqrt{\text{R}^2+{\text{X}^2_\text{L}}}=\sqrt{\text{R}^2+{(\omega\text{L})^2}}$

The current $\text{I}_\text{V}=\frac{\text{E}_\text{V}}{\text{Z}}$

$\frac{\text{E}_\text{v}}{\sqrt{\text{R}^2+\omega^2\text{L}^2}}$

2. 2 A

Explanation:

Since current lead and lag are same So, circuit is in resonance, so, circuit is purely resistive circuit.

So, $\text{i}=\frac{\text{V}}{\text{R}}$

$=\frac{200}{100}$

$=2\text{A}$

2. 0.7A

Explanation:

Resonant frequency, $\text{w}_\text{o}=\frac{1}{\sqrt{\text{LC}}}=\frac{1}{\sqrt{(10\times10^{-3})(10^{-6})}}=\frac{10^4\text{rad}}{\text{s}}$

New frequency $\omega=(0.9)\omega_\text{o}=9\times\frac{10^3\text{rad}}{\text{s}}$

We have new $\text{X}_\text{L}=\omega\text{L}=9\times10^3\times10\times10^{-3}=90\Omega\text{ and}\text{ X}_\text{C}=\frac{1}{\omega\text{C}}=111.11\text{ohm.}$

Thus we calculate new Z as

$\sqrt{3^2+[90-111.11]^2}$

$=21.32\Omega$

Current amplitude $=\frac{\text{V}_\text{o}}{\text{Z}}=\frac{15}{21.32}=0.7\text{A}$

1. high impedance

Explanation

$\text{Z}=\frac{1}{\text{WC}}$

$\text{W}\downarrow\text{Z}\uparrow$

3. time

Explanation:

$\text{f}=\frac{1}{\sqrt{\text{LC}}}$

$\text{or}{\sqrt{\text{LC}}}=\text{T}$

3. i₁ < i₂

Explanation:

$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$\text{i}=\frac{\text{dQ}}{\text{dt}}$

$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$\in_0\text{C}\times100\pi\Big[\sin\big(100\pi\text{s}^{-1}\big)\text{t}\Big]\\+\in_0\text{C}\times500\pi\Big[\sin\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$=100\text{C}\pi\in_0\cos\Big[\big(100\pi\text{s}^{-1}\big)\text{t}+\phi_1\Big]\\+500\text{C}\pi\in_0\cos\Big[\big(500\pi\text{s}^{-1}\big)\text{t}+\phi_2\Big]$

$\text{i}_1=100\pi\in_0\text{C}$ and $\text{i}_2=500\pi\in_0\text{C}$

$\text{i}_2>\text{i}_1$

1. Here, the power factor $\cos\phi\geq0,\text{P}\geq0$.
2. The driving force can give no energy to the oscillator (P = 0) in some cases.
3. The driving force cannot syphon out (P < 0) the energy out of oscillator.

Solution:

Key Concept: Power Factor.

1. It may be defined as cosine of the angle of lag of lead (i.e., $\cos\phi$).
2. It is also defined as the radio of resistance and impedange $\Big(\text{i.e.,} \frac{\text{R}}{\text{Z}}\Big)$.
3. $\text{The radio}=\frac{\text{True power}}{\text{Apparent power}}=\frac{\text{W}}{\text{VA}}=\frac{\text{kW}}{\text{kVA}}=\cos\phi$

In the given problem power transferred,

$\text{P}=\text{I}^2\text{Z}\cos\phi$

where I is the current, Z = Impedance and $\cos\phi$ is power factor

1. As power factor, $\cos\phi=\frac{\text{R}}{\text{Z}}$

where R > 0 and Z > 0

$\Rightarrow\ \cos\phi>0\Rightarrow\ \text{P}>0$

2. When $\phi=\frac{\pi}{2}$ (in case of L of C), P = 0.
3. From (a), it is clear that P < 0 is not possible.

4. In a pure inductive circuit emf will be in phase with the current

Explanation:

In pure inductive circuit, 

voltage leads by $\frac{\pi}{2}$ 

3. $\frac{\text{L}}{2}$

Explanation:

$\text{f}=\frac{1}{2\pi\sqrt{(\text{LC})}}$

when C is doubled, L should be halved so that resonant frequency remains unchanged.

3. LC will have to be decreased

Explanation:

$\text{Resonant frequency}=\frac{1}{\sqrt{\text{LC}}}$

$\text{LC}\downarrow\text{if }\omega_\text{r}\uparrow$

2. increases

Explanation:

Resonant frequency in the series RLC circuit $\text{v}_\text{r}=\frac{1}{2\pi\sqrt{\text{LC}}}$

$\Rightarrow\text{v}_\text{r}\propto\frac{1}{\sqrt{\text{C}}}$

Thus resonant frequency of the circuit increases if the value of C decreases.

4. $\frac{5}{\pi}$

Explanation:

Capacitive reactance $=\frac{1}{\omega\text{c}}$

$=\frac{1}{2\pi\text{fc}}$

$=\frac{1}{2\pi2\times10^3\times50\times10^{-6}}$

$=\frac{5}{\pi}\Omega$

2. 25.6V

Explanation:

In phasor, $\text{V}_\text{R}=20\angle0;\text{V}_\text{L}=16\angle(-90)$

Total potential difference is $=\text{V}_\text{R}+\text{V}_\text{L}=25.6\angle(-38.66)$

Magnitude of total potential difference = 25.6V

3. 1

Explanation:

At steady state inductor behave like short circuit.

$\text{i}=\frac{\text{v}}{\text{r}}=\frac{10}{10}=1\text{A}$