Physics M.C.Q [1M]

4. Zero

Explanation:

Power, $\text{P}=\text{I}_\text{rms}\times\text{V}_\text{rms}\times\cos\phi$

In the given problem, the phase difference between voltage and current is $\frac{\text{P}}{2}$ Hence

$\text{P}=\text{I}_\text{rms}\times\text{V}_\text{rms}\times\cos\big(\frac{\pi}{2}\big)=0$

4. (d)

Explanation:

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

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

$=2\pi\text{fL}-\frac{1}{2\pi\text{fC}}$

1. 50 Volt

Explanation:

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

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

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

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

Explanation:

Resonance frequency f of an L - C circuit can be written as

Resonance frequency $\text{f}=\frac{1}{2\pi\sqrt{\text{LC}}}$ where L = inductance and C is capacitance.

3. A.C. only

Explanation:

Capacitive reactance is given as $\text{X}_\text{C}=\frac{1}{\omega\text{C}}$

From this relation we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short

circuit. Likewise, as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC.

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.

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. be 4 times

Explanation:

inductive reactance $=2\pi\text{fL}$

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

1. Inductor and capacitor.

4. Resistor, inductor and capacitor.

Solution:

Compare the given circuit by predicting the variation in their reactances with frequency. So, that then we can decide the elements.

Reactance of an inductor of inductance L is $\text{X}_\text{L}=2\pi\text{vL}$, where v is the frequency of the AC circuit.

X_c = Reactance of the capacitive circuit

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

With an increase in frequency in (f) of an AC circuit, R remains constant, inductive reactance (X_L) increases and capacitive reactance (X_C) decreases. For an L-C-R circuit,

Z- Impedance of the circuit,

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

$=\sqrt{\text{R}^2+\bigg(2\pi\text{vL}-\frac{1}{2\pi\text{vC}}\bigg)^2}$

As frequency (v) increases, Z decreases and at certain value of the frequency known as resonant frequency (v₀), impedance Z is minimum that is Z_min = R current varies inversely with impedance and at Z_min current is maximum.

3. 90°

Explanation:

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

current leads voltage by $\frac{\pi}{2}$ i.e., current and voltage differ in phase by 90°

2. $\big(\frac{10}{\sqrt{2\text{V}}}\big)$

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. 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​.

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

Explanation:

Charge on the capacitor at any time `t' is

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

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

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

4. none of these

Explanation:

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

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

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

$=\frac{240}{30}$

$=8\text{A}$

1. 50V

Explanation:

$\text{V}=\sqrt{\text{V}_\text{R}^2+(\text{V}_\text{L}-\text{V}_\text{C})^2}=\sqrt{30^2+(60-20)^2}=50\text{V}$

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.

Explanation:

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

X_C​ and f are inversely proportional.

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

2. 0.5 A

Explanation:

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