Physics M.C.Q [1M]

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.

2. $10^4\Omega$

4. $155\Omega$

Explanation:

Inductance of the solenoid $=\text{L}=\frac{\mu\text{N}^2\text{A}}{1}=0.493\text{H}$

reactance of this solenoid $=\text{L}\omega=\text{L}2\pi\text{f}=155\Omega$

3. C, 9C

Explanation:

$\frac{\text{f}_\text{max}}{\text{f}_\text{max}}=3$

$\therefore\frac{\sqrt{\text{LC}_\text{max}}}{\sqrt{\text{LC}_\text{min}}}=3$

$\frac{\text{C}_\text{max}}{\text{C}_\text{min}}=9$

${\text{C}_\text{min}}=\text{C}$

$\therefore{\text{C}_\text{max}}=\text{9C}$

2. 25.6V

Explanation:

Potential difference across the circuit

$\text{V}=\sqrt{\text{V}^2_\text{R}+\text{V}^2_\text{L}}=\sqrt{20^2+16^2}=\sqrt{656}=25.6\text{V}$

3. Voltage is more but current is less

Explanation:

The power cables have some resistance. 

Power lost in the wires can be calculated as P = I²R with R as the resistance of the wires and I as the current that passes through them.

Power at the load is P = VI. 

From this one can see that if voltage is increased by say n times, then only $\frac{1}{\text{n}}$​ the current is required to deliver the same power. However, if $\frac{1}{\text{n}}$​ current is passed

on the same wires, only $\frac{1}{\text{n}^2}$ of the power will be lost.

3. $\text{I}^2\text{R}$

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

Explanation:

Resonant frequency, $\text{f}_\text{r}=\frac{1}{2\pi\sqrt{\text{LC}}}$

As the frequency is unchanged so f_r ​= f_r′​

LC = L′C ′= L′(2C)

$\text{L}=\frac{\text{L}}{2}$

2. inversely proportional to frequency

Explanation:

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

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

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

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

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

1. 200V

Explanation:

General sinusoidal voltage variation is given by:

$\text{V}=\text{V}_0\sin(\omega\text{t})$

Here, in this equation V₀​ is peak voltage.

So, on comparing both equation we get:

$\text{V}_0=200\text{V},\omega=314$

1. If R is increased current will decrease.

Explanation:

$\text{I}=\frac{\text{V}}{\text{Z}}=\frac{\text{V}}{\sqrt{\text{R}^2\Big(\omega\text{L }\sim}\frac{1}{\omega\text{C}}\Big)^2}$

By increasing R, current will definitely decrease by change in L or C, current may increase or decrease.

3. $\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$

Explanation:

$\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}$

$\text{I}_\text{rms}=\frac{\int\limits_0^\text{T}\text{I}^2\text{dt}}{\int\limits_0^\text{T}\text{dt}}$

if $\text{I}=\cos\omega\text{t}$

$\text{I}_\text{rms}^2=\frac{\text{I}_0^2}{2}$

$\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}$

Than $\text{i}_\text{rms}^2=\frac{\text{i}_1^2}{2}+\frac{\text{i}_2^2}{2}$

$\text{i}_\text{rms}=\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$

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

Explanation:

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

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

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

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

3. 120 V

Explanation:

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

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

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

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

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

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.

4. 4 times

Explanation:

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

1. $2\text{C}$

Explanation:

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

For f to be constant the product LC must be constant.

So, if we half the value of inductance then the value of capacitance must be doubled.

C should be changed to 2C.

1. DC.

Explanation:

$\text{X}_\text{C}\frac{1}{\omega\text{C}}=\frac{1}{0\times\text{C}}$ $\bigg\{\text{in}\stackrel{{\text{DC}}}{{\omega = 0 }}\bigg\}$

$=\infty$

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

Explanation:

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

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.

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 $\text{I}=\frac{\text{V}}{\text{Z}}$

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

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

1. 50 Hz

Explanation:

$\text{Y}=100\text{V, C}=2\mu\text{F}=20\times10^{-6}$

$\text{I}=0.628\text{A,v}=?$

$\text{X}_\text{C}=\frac{\text{V}}{\text{I}}=\frac{100}{0.628}$

$\Rightarrow\frac{1}{2\pi\text{vC}}=\frac{100}{0.628}$

$\text{v}=\frac{0.628}{100\times2\pi\text{C}}$

$=50\text{ Hz}$

4. $10^5\Omega$

Explanation:

$\text{Reactance}=\frac{1}{\text{C}\omega}\Omega$

$=\frac{1}{10^{-6}\times10}=\frac{1}{10^{-5}}$

$=10^5\Omega$

1. $\frac{1}{\sqrt{2}}\text{A}.$​

Solution:

Key concept: It decreases voltage and increases current

V_S < V_P

N_S > N_P

E_S < E_P

i_S > i_P

R_S < R_P

t_S > t_P

k < l

According to the problem output/secondary voltage V_S = 24V

Power associated with secondary P_S = 12W

$\text{I}_\text{S}=\frac{\text{P}_\text{S}}{\text{V}_\text{S}}=\frac{12}{24}=0.5\text{A}$

Amplitude of the current in the secondary winding

$\text{I}_0=\text{I}_\text{S}\sqrt{2}$

$=(0.5)(1.414)=0.707=\frac{1}{\sqrt{2}}\text{A}$

3. $141\Omega$

Explanation:

Total impedance = R + j(X_L​ - X_C​)

$=100+\text{j}100$

$=100\sqrt{2}\measuredangle45$

$=141\measuredangle45$

2. $5\text{mA}$

Explanation:

$\text{X}_\text{C}=\frac{1}{\text{C}\omega}=10000\Omega$

ammeater reading $=\text{I}_\text{rms}=\frac{\text{V}_\text{rms}}{\mid\text{jX}_\text{c}\mid}=\frac{50}{10000}=5\text{mA}$

1. effective resistance due to capacity

Explanation:

Capacitive reactance in an A.C circuit is: $\text{X}_\text{C}=\frac{1}{\omega\text{C}}\text{ohm}$

where, C is the capacitance of capacitor and $\omega=2\pi\text{rn}$ (n is the frequency of A.C source)