Physics M.C.Q (1 Marks)

3. 14.4W.

Solution:

According to the problem, X_L = 1Ω, R = 2Ω,

E_rms = 6V, P_av = ?

Average power dissipated in the circuit

$\text{P}_\text{av}=\text{E}_\text{rms}\text{I}_\text{rms}\cos\phi \ .....(\text{i})$

$\text{I}_\text{rms}=\frac{\text{E}_\text{rms}}{\text{Z}}$

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

$=\sqrt{4+1}=\sqrt{5}$

$\text{I}_\text{rms}=\frac{6}{\sqrt{5}}\text{A}$

$\cos\phi=\frac{\text{R}}{\text{Z}}=\frac{2}{\sqrt{5}}$

$\text{P}_\text{av}=6\times\frac{6}{\sqrt{5}}\times\frac{2}{\sqrt{5}}\ \ [\text{from Eq. (i)]}$

$=\frac{72}{\sqrt{5}\sqrt{5}}=\frac{72}{5}=14.4\text{W}$​​​​​​​

2. 2.828A

Explantion:

Given equation, $\text{e}=80\sin100\pi\text{t}.(\text{i})$

Standard equation of instantaneous voltage given by E =

$\text{e}_\text{m}\sin(\omega\text{t})......(\text{i})$

Compare (i) and (ii), we get e_m​ = 80V

where e_m​ is the voltage amplitude.

Current amplitude $\text{I}_\text{m}=\frac{\text{e}_\text{m}}{\text{Z}}$ where Z = impedence

$=\frac{80}{20}=4\text{A}$

$\text{I}_\text{r.m.s}=\frac{4}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}=2.828\text{A}$

2. $\frac{1}{2\pi\text{f}(2\pi\text{fL + R})}$

Explanation:

Here X_C​ - X_L ​= R

$\Rightarrow\frac{1}{2\pi\text{f}}=(\text{R}+2\pi\text{fL})$

$\Rightarrow\text{C}=\frac{1}{2\pi\text{f}(2\pi\text{fL + R})}$

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

Explanation:

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

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

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

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

4. $\text{None of these}$

Explanation:

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

1. 100V

Explanation:

The power supplied to the resistor by DC source $=\frac{\text{V}^2_\text{dc}}{\text{R}}$

Energy given by AC source $\int_{0}^{\text{T}}=\frac{\text{V}^2_0}{\text{R}}\text{dt}$

Hence, $\int_{0}^{\text{T}}=\frac{\text{V}^2_0\sin^2\omega\text{t}}{\text{R}}\text{dt}=\frac{1}{2}\times\frac{\text{V}^2_\text{dc}\text{T}}{\text{R}}$

$\Rightarrow\text{V}_0=\text{V}_\text{dc}=100\text{V}$

1. $2.2\text{mA}$

Explanation:

Effective current is the rms value. Here, 220V is the labelled value of AC which is also the rms value. Hence,

$\text{I}_\text{rms}=\frac{\text{E}_\text{rms}}{\text{R}}$

$\text{I}_\text{rms}=\frac{220}{100\times10^3}$

$\text{I}_\text{rms}=2.2{\text{mA}}$

2. the current and P.d across the resistance lags behind the P.d. across the inductance by angle $\frac{\pi}{2}$

Explanation:

This is very fundamental. If we apply separate voltages across resistance and inductor, then in resistance, current and voltage both are in same phase whereas in inductor, current across it lags p.d across it by $\frac{\pi}{2}.$
Now, when we apply voltage across inductor and resistance connencted in series then current through both of them will be same because of KCL. therefore voltage across resistor will be in same phase with current whereas voltage across inductor will lead the current across it by $\frac{\pi}{2}.$ therefore current and voltage across resistor lags voltage across inductor by $\frac{\pi}{2}$

4. none of these

Explanation:

$\sin\phi=\frac{\text{X}}{\text{Z}}=\frac{1}{\sqrt{3}}$

$\therefore\phi=\sin^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

2. May be zero.

Explanation:

1. $\text{V}=\text{V}_0\sin\omega\text{t}$

$\omega=2\pi\text{f}=2\times3.14\times50$

$\omega=314$

$\text{V}_\text{avg}=\frac{\int\limits_0^{0.01}\text{V}\text{dt}}{\int\limits_0^{0.01}\text{dt}}$

$=\text{V}_0\Big(\frac{1\cos\omega\text{t}}{\omega}\Big)_0^{0.01}$

$=\frac{\text{V}_0}{\omega\times0.01}\big(1-\cos\omega(0.1)\big)$

$=\frac{\text{V}_0}{314\times0.01}\big(1-\cos(314\times0.01)\big)$

$=\frac{\text{V}_0}{3.14}\big(1-\cos(314)\big)$

$=\frac{\text{V}_0}{3.14}\big(1-\cos\pi\big)$

$=\frac{2\text{V}_0}{\pi}=140.127\text{volt}$

if $\text{V}=\text{V}_0\cos\omega\text{t}$

$\text{V}_\text{avg}=\frac{\int\text{V d}\rho}{\int\text{dt}}=0$

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.

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. 1.0012

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. $-\frac{\pi}{2}$

Explanation:

In a purely capacitive circuit, current leads the voltage by $-\frac{\pi}{2}$

2. $\text{Between }50\Omega \text{ to } 70\Omega$

Explanation:

Characteristic impedance of a coaxial cable is $\text{Between }50\Omega \text{ to } 70\Omega$

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

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

1. 20V, 10V, 10V

Explanation:

Circuit is at resonance (VL = VC)

$\therefore$ circuit is purely resistance

Resistance is doubled, current in the circuit is half the initial value

$\therefore$ New current $\text{I′}=\frac{\text{I}}{2}$

$\therefore$ VR = 20V (equal to applied voltage earlier)

 VL = 10V

 VC = 10V

1. be 4 times

Explanation:

inductive reactance $=2\pi\text{fL}$

therefore when f is made 4 times, inductive reactance also becomes 4 times

2. $5\sqrt{\frac{3}{2}}\text{A}$

Solution:

Key concept: Equation for i and V: Alternating current or voltage varying as sine function can be written as

$\text{i}=\text{i}_0\ \sin\omega\text{t}=\text{i}_0\ \sin2\pi\text{v t}=\text{i}_0\sin\frac{2\pi}{\text{T}}\text{t}$

and $\text{V}=\text{V}_0\ \sin\omega\text{t}=\text{V}_0\ \sin2\pi\text{v t}=\text{V}_0\sin\frac{2\pi}{\text{T}}\text{t}$

where i and V are instantaneous values of current and voltage,

i₀ and V₀ are peak values of current and voltage

$\omega$ = Angular frequency in $\frac{\text{red}}{\text{sec}}$, v = Frequency in Hz and T = time period

Accoeding to the problem, f = 50Hz, I_rms = 5A

$\text{t}=\frac{\text{I}}{300}\text{s}$

$\text{I}_0=\text{Peak value}=\sqrt{2}(\text{I}_\text{rms})=5\sqrt{2}$

$=5\sqrt{2}\text{A}$

From, $\text{I}=\text{I}_0\sin\omega\text{t}=5\sqrt{2}\sin2\pi\text{vt}=5\sqrt{2}\sin2\pi\times50\times\frac{1}{300}$

$=5\sqrt{2}\sin\frac{\pi}{3}=5\sqrt{2}\times\frac{\sqrt{3}}{2}=5\sqrt{\frac{3}{2}}\text{A}$

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)

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

3. 1

Explanation:

At steady state inductor behave like short circuit.

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

2. 2 A

Explanation:

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

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

$=\frac{200}{100}$

$=2\text{A}$

1. Current.
2. Induced emf in the inductor.

Explanation:

$\text{I}=\text{I}_0\sin\omega\text{t}$

Average value of current over a cycle = 0

$\text{V}=\text{V}_0\cos\omega\text{t}$

$=\text{V}_0\sin\bigg(\omega\text{t}+{\frac{\pi}{2}}\bigg)$

Average value of induced emf in inductor over a cycle = 0.

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. Resistor and a capacitor.
4. Only a capacitor.

Solution:

This is the similar problem as we discussed above. In this problem, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreased as increase in frequency. So, one element that must be connected is capacitor. We can also connect a resistor in series.

For a capacitive circuit,

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

When frequency increases, X_C decreases. Hence current in the circuit increases.

Importance point: Resistive, Capacitive Circuit (RC-Circuit)

V_R = iR,

V_C = iX_C,

V_R = iR

1. Applied voltage: $\text{V}=\sqrt{\text{V}^2_\text{R}+\text{V}^2_\text{C}}$

2. Impedance: $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_\text{C}}=\sqrt{\text{R}^2+\Big(\frac{1}{\omega\text{C}}\Big)^2}$

3. Current: $\text{i}=\text{i}_0\sin(\omega\text{t}+\phi)$

4. Peak current: $\text{i}_0=\frac{\text{V}_0}{\text{Z}}=\frac{\text{V}_0}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}=\frac{\text{V}_0}{\sqrt{\text{R}^2+\frac{1}{4\pi^2\text{v}^2\text{C}^2}}}$

5. Phase difference: $\phi=\tan^{-1}\frac{\text{X}_\text{C}}{\text{R}}=\tan^{-1}\frac{1}{\omega\text{CR}}$

6. Power factor: $\cos\phi=\frac{\text{R}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$

7. Leading quantity: Current.

3. 4.6mA

Explanation:

Voltage of the source V = 20 volts

$\text{Z}=4.33\text{k}\Omega=4.33\times10^3\Omega$

Thus current in the circuit $\text{I}=\frac{\text{V}}{\text{Z}}$

$\therefore\text{I}=\frac{20}{4.33\times10^3}=4.6\times10^{-3}\text{A}$

$\Rightarrow\text{I}=4.6\text{mA}$

3. time

Explanation:

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

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

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.

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. Inductive

Explanation:

At resonant frequency

$\text{X}_\text{L}=\text{X}_\text{C}=\Big(\omega\text{L}=\frac{1}{\omega\text{C}}\Big)$

At frequencies higher than resonance frequencies

$\text{X}_\text{L}>\text{X}_\text{C}$

i.e., behaviour is inductive.