Physics M.C.Q (1 Marks)

Explanation:

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

X_C​ and f are inversely proportional.

3. I = 0.5 A

Explantion:

The current in inductor and capacitor is always at an phase difference of 180°

for V $=\text{V}_\text{o}\sin\omega\text{t}$

Capacitor, $\text{i}=\text{i}_\text{o}\sin(\omega\text{t}-\frac{\pi}{2})$

Capacitor, $\text{i}=\text{i}_\text{o}\sin(\omega\text{t}+\frac{\pi}{2})$

So, the current from both branches will be 0.9 + (-0.4) = 0.5A

1. 0 and 2A

Explanation:

In the problem $\text{XC}=4\Omega$ and $\text{XL}=4\Omega$

So, V across XC​ and XL​ will be same and in opposite direction, So net voltage will be zero. Since voltmeter is connected parallel to Capacitor and inductor so, it will read 0 volts.

Current $=\frac{\text{V}}{\text{impedance}}$

Z = R as X_L ​= X_C​

Current $=\frac{90}{45}=2\text{A}$

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.

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. B₂

Explanation:

Impedance in branch containing bulb1

$\text{Z}_1=200\Omega$

impedance in branch containing bulb2

$\text{Z}_2\sqrt{\text{R}^2+\text{X}_\text{L}}^2$

$\text{Z}_2=\sqrt{100^2+100^2}$

$\text{Z}_2=100\sqrt{2}$

Since,

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

B₂​ will glow more than B₁​.

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. 0 A and 0 A

Explanation:

$\text{I}=\frac{220}{25+\text{j}\Big(0.1\times2\pi\text{f }-\frac{1}{10^{-5}2\pi\text{f}}\Big)}$

$\text{I}=0\text{ at }\text{f}=0$

$\text{I}=0\text{ at }\text{f}=\infty$

2. $40\sqrt{2}\text{V}$

Explanation:

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

$=\sqrt{40^2+(40-80)^2}$

$=\sqrt{40^2+40^2}$

$=40\sqrt{2}\text{V}$

2. $\text{At t}=2\tau,\text{q}={\text{CV}}(1-\text{e}^{-2})$

Explanation:

Charge on the capacitor at any time `t' is

$\text{q}=\text{CV}(1-\text{e}\frac{-\text{t}}{\tau})$

$\text{at }\text{t}=2\tau$

$\text{q}=\text{CV}(1-\text{e}^{-2})$

3. 6kHz

Explanation:

Frequency of applied voltage and voltage or current accross any component are same. As we know,

$\text{I}=\frac{\text{V}}{\text{Z}}$ and Z has nothing to do with frequency 

$\Rightarrow $$ \text{IandV}$$$ have same frequency. 

And also according to KVL V = V₁ ​+ V₂​ + V_3​

⟹ voltage and current across any component have same frequency i.e 6kHz.

1. 0 V, 8 A

Explanation:

Voltmeter reading is zero since voltage in both Capacitor and inductor is same in magnitude but opposite in sign because current in series is same and both have same reactance.

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

$=\frac{240}{30}$

$=8\text{A}$

3. 120 V

Explanation:

At resonance the voltage across L and C are same but opposite.

So, at resonance $\mid\text{V}_\text{L}\mid=\mid\text{V}_\text{C}\mid$

$\therefore\text{V}_\text{R}=\text{V}_\text{app}=120\text{V}$

2. decrease

Explanation:

In complex plane $\text{V}=\frac{\text{V}_\text{AC}}{1+\text{jRC}\omega}$

Therefore as $\omega$(frequency) increases, V decreases.

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

Explanation:

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

Ceq ​= C + 2C = 3C

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

2. 0.5 A

Explanation:

Wattless current $=\text{I}_\text{max}\sin(\tan^{-1}[\frac{\text{L}\omega}{\text{R}}])=\frac{220}{\sqrt{\text{R}^2+\text{L}^2\omega^2}}\sin(\tan^{-1}[\frac{\text{L}\omega}{\text{R}})=0.5\text{A}$

Explanation:

Capacitor reactance is given by: $\text{X}_\text{c}=\frac{1}{2\pi\text{fC}}$ 

C is the capacitance. 

X_c​ & f are inversely proportional.

2. increases directly with frequency

Explanation:

The inductive reactance $\text{Xl}=\omega\text{L}$

Hence, $\text{X}_1\propto\omega$

As frequency increases $\rightarrow\omega$

Therefore, inductive reactance increases with frequency.

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

3. The meter reads not v but < v² > and is calibrated to read $\sqrt{<\text{v}^2>}$.​​​​​​​

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. 50 Volt

Explanation:

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

$\text{V}=\sqrt{30^2+(80-40)^2}$

$\text{V}=50\text{ Volt}$

4. 4 times

Explanation:

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

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. 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. inversely proportional to frequency

Explanation:

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

$\therefore\text{X}_\text{C}\propto\frac{1}{\omega}$

3. 16mH

Explanation:

$\tan(45)=1\frac{\text{L}\omega}{\text{R}}$

$\text{L}=\frac{\text{R}}{\omega}=\frac{\text{R}}{(2\pi1000)}=.016\text{H}$

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:

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

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

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

$20\Omega$

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

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 $\frac{\pi}{2}$