Physics M.C.Q [1M]

3. 311V

Explanation:

The rated voltage in bulb is rms voltage.

$\text{V}_\text{rms}=\frac{\text{V}_0}{\sqrt{2}}$

$\text{V}_0=\sqrt{2}\times220=311.08\text{V}$

3. $20\mu\text{F}$

4. None of the above.

Explanation:

$\text{V}=\sqrt{\text{V}_\text{R}^2+(\text{V}_\text{C}-\text{V}_\text{L})^2}=10\text{V}$

V_C​ > V_L​, hence current leads the voltage.

Power factor $=\cos\phi=\frac{8}{10}=0.8$

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.

1. Zero average current.

4. Voltage and current possibly differing in phase $\phi$ such that $\Big|\phi\Big|<\frac{\pi}{2}$.

Solution:

Alternation currents are used for household supplies, which are having zero average value over a cycle.

The line is having some resistance, so power factor $\cos\phi=\frac{\text{R}}{\text{Z}}\neq0$

So, $\phi\neq\frac{\pi}{2}\Rightarrow\ \phi<\frac{2}{\pi}$

i.e., phase lies between 0 and $\frac{\pi}{2}$.

Important point: The average value of alternating quantity for one complete cycle is zero.

The average value of ac over half cycle $\Big(\text{t}=0\text{ to }\frac{\text{T}}{2}\Big)$

$\text{i}_\text{av}=\frac{\int_0^{\frac{\text{T}}{2}}\text{idt}}{\int_0^{\frac{\text{T}}{2}}\text{dt}}=0.637\text{i}_0=63.7\% \ \text{of i}_0$

Similarly $\text{V}_\text{av}=\frac{2\text{V}_0}{\pi}=0.637\text{V}_0=63.7\%\ \text{of V}_0.$

1. lags the applied voltage

Explanation:

Capacitive reactance is given by. $\text{X}_\text{C}=\frac{1}{\text{wC}}$

Inductive reactance is given by $\text{X}_\text{L}={\text{wL}}$

At resonance, $\text{X}_\text{L}={\text{X}_\text{C}}\Rightarrow \text{w}\text{L}=\frac{1}{\text{wC}}$

But a frequency higher than resonance frequency, X_L​ > X_C​

So the circuit behaves as a inductive circuit at a frequency higher that resonant frequency and the current lags behind the voltage in an inductive circuit.

3. Maximum but finite

Explanation:

At resonance X_L​ = X_C​

⇒ R & current is maximum but finite, which is

$\text{I}_\text{max}=\frac{\text{E}}{\text{R}}$ where E is applied voltag.

2. $\frac{1}{6\pi\sqrt{\text{LC}}}$

Explanation:

Equivalent inductance Leq ​= L + 2L = 3L

Ceq ​= C + 2C = 3C

$\therefore$ Frequency of oscillation $\text{f}=\frac{1}{2\pi\sqrt{\text{L}_\text{eq}\text{C}_\text{eq}}}=\frac{1}{6\pi\sqrt{\text{LC}}}$

3. $30\pi$

3. About 310V.

Explanation:

$\text{V}_\text{rms}=220\text{V}$

$\text{V}_\text{p}=\sqrt{2}\times\text{V}_\text{rms}$

$=220\times1.414=311\text{volt}$

2. 16 V

Explanation:

$\text{V}^2={\text{V}^2_\text{R}}+{\text{V}^2_\text{c}}$

$20^2=12^2+\text{V}^2_\text{c}$

$\text{V}_\text{c}=\sqrt{20^2-12^2}$

$\text{V}_\text{c}=16\text{ V}$

1. $\frac{1}{100}\text{second}$

Explanation:

In India, the frequency of AC voltage is 50 Hz.

It means 50 waves will be produced in 1 s.

In one wave, the direction is changed 2 times.

Thus, in 50 waves, the direction will be altered 50 × 2 = 100 times.

i.e. the direction is changed 100 times in 1 s.

Thus the direction is changed in every $\frac{1}{100}\text{second}$

2. Another capacitor should be added in parallel to the first.

Solution:

Key Concept: Resonant frequency (Natural frequency)

At resonance $\text{X}_\text{L}=\text{X}_\text{C}\Rightarrow\ \omega_0\text{L}=\frac{1}{\omega_0\text{C}}$

$\Rightarrow\ \omega_0=\frac{1}{\sqrt{\text{LC}}}\frac{\text{red}}{\sec}$

$\Rightarrow\ \text{v}_0=\frac{1}{2\pi\sqrt{\text{LC}}}\text{Hz}$

Resonant frequency in an L-C-R circuit is given by

$\text{v}_0=\frac{1}{2\pi\sqrt{\text{LC}}}$

If L or C increases, the resonant frequency will reduce.

To increase capacitance, we must connect another capacitor parallel to the first.

1. 10.2

Explanation:

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

$=\sqrt{(10)^2+(8-6)^2}$

$=\sqrt{10^2+2^2}$

$=\sqrt{100+4}$

$=10.2\Omega$

2. Capacitive reactance is inversely proportional to the frequency of the current

Explanation:

Capacitative reactance is an opposition to the change of voltage across an element.

It is denoted by XC​ and is inversely proportional to the signal frequency (f) and the capacitance C

$\text{X}_\text{c}=\frac{1}{2\pi\text{fc}}$

2. $\text{R}$

Explanation:

At resonance $\frac{1}{\omega\text{C}}=\omega\text{L}$

$\therefore$ impedance = R

1. Decrease to one-half of the original value

Explanation:

Resonant frequency in series LCR circuit: $\omega=\sqrt{\frac{1}{\text{LC}}}$

Resonant frequency in series LCR circuit, $\omega=\sqrt{\frac{1}{\text{LC}}}=\sqrt{\frac{1}{\text{2L}\times\text{2C}}}=\frac{\omega}{2}$

3. V_a​ > V_b​

Explanation:

For series C - R circuit, the impedance $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_\text{C}}$ where $\text{X}_\text{C}=\frac{\text{i}}{\omega\text{C}}$ and current $\text{I}=\frac{\text{V}}{\text{Z}}$

When the capacitor is filled by mica, the capacitance will be increased. If C increases, X_C​ decreases, so the current will increase and 

hence voltage across resistance increases and voltage across capacitor decreases. thus, V_a​ > V_b​

3. The charge on the plates is in phase with the applied voltage.
4. Power delivered to the capacitor is zero.

Solution:

If the alternating voltage is applied to the capacitor, the plate connected to the positive terminal of the source will be at higher potential and the plate connected to the negative terminal of source will be at lower potential. So the plates capacitor is charged.

If $\text{V}=\text{V}_0\sin\omega\text{t},\text{Q}=\text{C V}_0\sin\omega\text{t}$ or we can say that Q and emf are in pahse.

As $\text{P}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\phi$ and in case of a capacitor, $\phi=\frac{\pi}{2}\text{P} = 0,$ or we can say that power delivered to the capacitor is zero.

⇒ P_av = power delivered = 0.

1. A is true but B is false

Explanation:

$\text{Z}_\text{L}=\text{WL}\ \ \ \text{Z}_\text{C}=\frac{1}{\text{WC}}$

$\text{w}\uparrow,\text{Z}_\text{L}\uparrow\text{Z}_\text{c}\downarrow$

3. $20\Omega$

Explanation:

$\text{impedance}×\cos\theta = \text{resistance}$

$\text{impedance} = \frac{\text{resistance}}{\cos\theta}$

$=\frac{\sqrt{300}}{\frac{\cos\pi}{6}}$

$20\Omega$

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

Explanation:

In the above image waveform of current and voltage in puerly inductive circuit with time is shown.

It is clear from the image that current lags voltage by 90°.

Hence phase angle between current and voltage in purely inductive circuit is $\frac{\pi}{2}$

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

Explanation:

Current leads voltage by $\frac{\pi}{2}$

$\therefore$ phase difference is $=\frac{\pi}{2}$

3. 50V

Explanation:

Supply voltage of an LCR circuit

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

since inductor and capacitor potentials are out of phase with each other

$=\sqrt{40^2+(60-30)^2\text{V}}$

$=50\text{V}$

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. 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. 0.2A

Explanation:

I_C​ is 90° ahead of the applied voltage and I_L​ lags behind the applied voltage by 90°. So, there is a phase difference of 180° between IL​ and IC​

$\therefore$ I = I_C​ - I_L ​= 0.2A

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. $\text{None of these}$

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

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

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 $\frac{\pi}{2}$