Physics 4 Marks Questions

1. (c) Power

Explanation:

In an ideal transformer, there is no power loss. The efficiency of an ideal transformer is $\eta=1$ (i.e 100%) i.e. input power = output power.

2. (d) Obtain desired ac voltage and current.

Explanation:

Transformer is used to obtain desired ac voltage and current.

3. (c) ac 110V

Explanation:

For a transformer, $\frac{\text{V}_\text{S}}{\text{V}_\text{p}}=\frac{\text{n}_\text{S}}{\text{n}_\text{p}}$

Where N denotes number of turns and V = voltage.

$\therefore{\text{V}_\text{S}}={\text{ac}\ 110}{\text{V}}$

4. (a) Current through its secondary is about four times that of the current through its primary.

Explanation:

In a transformer the primary and secondary currents are related by,

$\text{I}_\text{S}=\Big(\frac{\text{N}_\text{S}}{\text{N}_\text{P}}\Big)\text{I}_\text{P}$

And the Voltage are related by,

$\text{V}_\text{S}=\Big(\frac{\text{N}_\text{P}}{\text{N}_\text{S}}\Big)\text{V}_\text{P}$

where subscripts p and s refer to the primary and secondary of the transformer.

Here, $\text{V}_\text{P}=\text{V},\frac{\text{N}_\text{P}}{\text{N}_\text{S}}=4\ \therefore\text{I}_\text{P}=4\text{I}_\text{P}$

and, $\text{V}_\text{S}=\Big(\frac{1}{4}\Big)\text{V}=\frac{\text{V}}{4}$

5. (c) 90%

Explanation:

The efficiency of the transformer is:

$\eta=\frac{\text{output power}(\text{p}_\text{out})}{\text{intput power}(\text{p}_\text{in})}\times100$

Here, P_out = 100W, P_in = (220V)(0.5A) = 110W

$\therefore\eta=\frac{100\text{W}}{110\text{W}}\times100=90\%$

1. (d) 1J

Explanation:

Energy, $\text{E}=\frac{1}{2}\frac{\text{Q}^2}{\text{C}}=\frac{(10\times10^{-3})^2}{2\times50\times10^{-6}}=1\text{J}$

2. (a) 159.24Hz

Explanation:

Frequency, $\upsilon=\frac{1}{2\pi\sqrt{\text{LC}}}$

$\upsilon=\frac{1}{2\pi\sqrt{20\times10^{-3}\times50\times10^{-6}}}$

$=\frac{10^3}{2\pi}=159.24\text{Hz}$

3. (d) $0,\frac{\text{T}}{2},{\text{T}},\frac{\text{3T}}{2}$

Explanation:

Total time period, $\text{T}=\frac{1}{\upsilon}=\frac{1}{159.24}=6.20\text{ms}$

Total charge on capacitor at time t, $\text{Q}'=\text{Q}\cos\frac{2\pi}{\text{T}}\text{t}$

For energy stored is electrical, we can write Q' = ± Q.

Hence, energy stored in the capacitor is completely electrical at, $\text{t}=0,\frac{\text{T}}{2},\text{t},\frac{3\text{T}}{2},\ ......$

4. (d) $\frac{\text{T}}{4},\frac{\text{3T}}{4},\frac{\text{5T}}{4}$

Explanation:

Magnetic energy is maximum when electrical energy is equal to zero.

Hence, $\text{t}=\frac{\text{T}}{4},\frac{\text{3T}}{4},\frac{\text{5T}}{4},\ ....$

5. (a) $20\Omega$

Explanation:

$\text{X}_\text{L}=\omega_\text{L}=2\pi\upsilon\text{L}$

$2 × 3.14 × 159.24 × 20 × 10^{-3}$

$\Rightarrow\text{X}_\text{L}=20\Omega$

1. (c) $8.8\Omega$

Explanation:

Inductive reactance,

$\text{X}_\text{L}= \text{W}_\text{L}=2\pi\mu\text{L}$

$=2\pi\times100\times14\times10^{-3}$

$\text{X}_\text{L}=8.8\Omega$

2. (b) Inductor
3. (b) 90º

Explanation:

In an inductor vol tag: leads the current by $\frac{\pi}{2}$ or current lags the voltage by $\frac{\pi}{2}.$

4. (b) 1H

Explanation:

The current in the inductor coil is given by,

$\text{I}_0=\frac{\text{E}_0}{\text{X}_\text{L}}=\frac{\sqrt{2}\text{E}_\text{V}}{2\pi\mu\text{L}}$

$\text{L}=\frac{\sqrt{2}\text{E}_\text{V}}{2\pi\mu\text{I}_0}=\frac{1.414\times200}{2\times3.14\times50\times0.9}=1\text{H}$

5. (a) 0.337A

Explanation:

Inductive reactance,

$\text{X}_\text{L}= \text{W}_\text{L}=2\pi\mu\text{L}$

$=2\pi\times3.14\times50\times2=628\Omega$

$\text{I}_0=\frac{\text{E}_0}{\text{X}_\text{L}}$

$\Rightarrow\text{I}_0=\frac{\sqrt{2}\times\text{E}_\text{V}}{\text{X}_\text{L}}=\frac{\sqrt{2}\times150}{628}=0.337\text{A}$

1. $\text{E}_\text{rms}=\frac{\text{E}_0}{\sqrt{2}}$

$\Rightarrow\ \text{E}_0=\text{E}_\text{rms}\sqrt{2}$

$=\sqrt{2}\times220=1.414\times220$

$=311.08\text{V}=311\text{V}$

2. Time taken for the current to reach the peak value = Time taken to reach the 0 value from r.m.s.

$\text{I}=\frac{\text{I}_0}{\sqrt{2}}$

$\Rightarrow\ \frac{\text{I}_0}{\sqrt{2}}=\text{I}_0\sin\omega\text{t}$

$\Rightarrow\ \omega\text{t}=\frac{\pi}{4}$

$\Rightarrow\ \text{t}=\frac{\pi}{4\omega}=\frac{\pi}{4\times22\pi\text{f}}$

$=\frac{\pi}{8\pi50}=\frac{1}{400}=2.5\text{ms}$

1. (d) $15\Omega$

Explanation:

Resistance of the two wire lines carrying power $=0.5\frac{\Omega}{\text{Km}}$

Total resistance $=(15+15)0.5=15\Omega$

2. (c) 600kW

Explanation:

Line power loss = I²R

RMS current in the coil,

$\text{I}=\frac{\text{P}}{\text{V}_1}=\frac{800\times10^3}{4000}=200\text{A}$

$\therefore$ Power loss = (200)² × 15 = 600kW

3. (d) 1400kW

Explanation:

Assuming that the power loss is negligible due to the leakage of the current.

The total power supplied by the plant,

= 800kW + 600kW = 1400kW

4. (b) 3000V

Explanation:

Voltage drop in the power line = IR

= 200 × 15 = 3000V

5. (d) 7000V

Explanation:

Total voltage transmitted from the plant,

= 3000V + 4000V = 7000V

1. (b) 11.63A

Explanation:

Inductance, L = 80mH = 80 × 10^-3H

Capacitance, $\text{C}=60\mu\text{F}=60\times10^{-6}\text{F},\ \text{V}=230\text{V}$

Frequency, $\upsilon=50\text{Hz}$

$\omega=2\pi\upsilon=100\pi\ \text{red}\ \text{s}^{-1}$

Peak voltage, $\text{V}_0=\text{V}\sqrt{2}=230\sqrt{2}\text{V}$

 Maximum current is given by, $\text{I}_0=\frac{\text{V}_0}{\Big(\omega\text{L}-\frac{1}{\omega\text{C}}\Big)}$

$\text{I}_0=\frac{230\sqrt{2}}{\Big(100\pi\times80\times10^{-3}-\frac{1}{100\pi\times60\times10^{-6}}\Big)}$

$\text{I}_0=\frac{230\sqrt{2}}{\Big(8\pi-\frac{1000}{6\pi}\Big)}=-11.63\text{A}$

Amplitude of maximum current, I₀ = 11.63A

2. (c) 8.23A

Explanation:

rms value of current,

$\text{I}=\frac{\text{I}_0}{\sqrt{2}}=-\frac{-11.63}{\sqrt{2}}=-8.23\text{A}$

Negative sign appears as $\text{W}_\text{L}<\frac{1}{\text{W}_\text{C}}.$

3. (a) Zero

Explanation:

Average power consumed by the inductor is zero because of phase difference of $\frac{\pi}{2}$ between voltage and current through inductor.

4. (b) Zero

Explanation:

Average power consumed by the capacitor is zero because of phase difference of $\frac{\pi}{2}$ between voltage and current through capacitor.

5. (a) Zero

1. (c) 17.42W

Explanation:

Average power transferred per cycle to resistance is $\text{P}_\text{V} = \text{I}^2_\text{V}\text{R}$ 

As, $\text{X}_\text{L}=\omega_\text{L}=2\pi\upsilon\text{L}=2\times\frac{22}{7}\times60\times0.6=226.28\Omega$

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

$=\frac{1}{2\times\frac{22}{7}\times60\times50\times10^{-6}}=53.03\Omega$

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

$=\sqrt{\text{10}^2+(\text{226.28}_-\text{53.03})^2}=173.53\Omega$

$\text{I}_\text{V}=\frac{\text{E}_\text{V}}{\text{R}}=\frac{230}{173.53}=1.32\text{A}$

$\text{P}_\text{V} = \text{I}^2_\text{V}\text{R}=(1.32)^2\times10=17.42\text{W}$

2. (a) Zero

Explanation:

$\text{P}_\text{V}=\text{E}_\text{V}\text{I}_\text{V}\cos\phi.$

In a capacitor, phase difference, $\phi=90^\circ$

$\therefore​​​​\text{P}_\text{L}=\text{E}_\text{V}\text{I}_\text{V}\cos90^\circ=\text{Zero}$ 

3. (d) Zero

Explanation:

$\text{P}_\text{L}=\text{E}_\text{V}\text{I}_\text{V}\cos\phi.$

In an inductor, phase difference, $\phi=90^\circ$

$\therefore​​​​\text{P}_\text{L}=\text{E}_\text{V}\text{I}_\text{V}\cos90^\circ=\text{Zero}$

4. (a) 17.42W

Explanation:

Total power absorbed per cycle 

P = P_R + P_C + P_L = 17.42 + 0 + 0 = 17.42W

5. (d) 6.2 × 10⁴ Joule

Explanation:

Energy spent = power × time

= 17.42 × 60 × 60

= 6.2 × 10⁴ Joule

1. (c) Ohm

Solution:

Ohm is the unit of capacitive reactance.

2. (a) $0.032\Omega$

Solution:

Capacitive reactance, $\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\mu\text{C}}$

$=\frac{1}{2\pi\times10^6\times5\times10^{-6}}$

$=0.032\Omega$

3. (b) Capacitor

Solution:

In capacitive circuit, resistance to the flow of current is offered by the capacitor.

4. (b) 90º
5. (a) 6.28 × 10^-2A

Solution:

Current, $\text{l}_\text{v}=\frac{\text{E}_\text{V}}{\text{X}_\text{C}}=\frac{1}{\frac{1}{2\pi\mu\text{C}}}=(2\pi\mu\text{C})\text{E}_\text{V}$

$\text{I}_\text{v}=2\times3.14\times50\times10^{-6}\times200$

$=6.28 × 10^{-2}\text{A}$

1. (c) 663.48Hz

Explanation:

Here, L = 0.12H, e = 480nF = 480 × 10^-9F

$\text{R} = 23\Omega,$ V = 230V

$\text{V}_0=\sqrt{2}\times230=325.22\text{V}$

$\text{I}_0=\frac{\text{V}_0}{\sqrt{\text{R}^2+\Big(\omega\text{L}-\frac{1}{\omega\text{C}}\Big)^2}}$

At resonance, $\omega\text{L}-\frac{1}{\omega\text{C}}=0$

$\omega=\frac{1}{\sqrt{\text{LC}}}=\frac{1}{\sqrt{0.12\times480\times10^9}}=4166.67\ \text{rad}\ \text{s}^{-1}$

$\upsilon_\text{R}=\frac{4166.67}{2\times3.14}=663.48\text{H}_\text{z}$

2. (a) 14.14A

Explanation:

Current, $\text{I}_0=\frac{\text{V}_0}{\text{R}}=\frac{325.22}{23}=14.14\text{A}$

3. (b) 2299.3W

Explanation:

Maximum power, $\text{P}_\text{max}=\frac{1}{2}(\text{I}_0)^2\text{R}$

$=\frac{1}{2}\times(14.14)^2\times23=2299.3\text{W}$

4. (d) 21.74

Explanation:

Quality factor, $\text{Q}=\frac{\text{X}_\text{R}}{\text{R}}=\frac{\omega_\text{r}\text{L}}{\text{R}}$

$=\frac{4166.67\times0.12}{23}=21.74$

5. (b) Current

1. (c) 3V

Explanation:

As, $\frac{\text{E}_\text{s}}{\text{E}_\text{p}}=\frac{\text{n}_\text{s}}{\text{n}_\text{p}}$

$\Rightarrow\text{E}_\text{s}=\text{E}_\text{p}\times\frac{\text{n}_\text{p}}{\text{n}_\text{p}}$

$=\frac{120\times50}{2000}=3\text{V}$

2. (d) 5A

Explanation:

$\text{I}_0=\frac{\text{E}_\text{S}}{\text{R}}\Rightarrow\text{I}_0=\frac{3}{0.6}=5\text{A}$

3. (a) 0.125A

Explanation:

As, $\frac{\text{I}_\text{P}}{\text{I}_\text{S}}=\frac{\text{E}_\text{S}}{\text{E}_\text{P}}$

$\Rightarrow\text{I}_\text{p}=\frac{\text{E}_\text{S}}{\text{E}_\text{P}}\times\text{I}_\text{S}=\frac{3}{120}\times5=0.125\text{A}$

4. (d) 15W

Explanation:

Power in primary, P_P = E_P × I_P = 120 × 0.125

= 15W

5. (a) 15W

Explanation:

Power in secondary coil, P_S = E_S × I_S = 3 × 5

= 15W

1. (b) $31.4\Omega$

Explanation:

Given, $\text{R}=12\Omega,\ \text{X}_C=14\Omega,\ \text{L}=0.1\text{H}$

$\text{X}_L=\omega_\text{L}=2\pi\upsilon\text{L}$

$= 2\times3.14\times50\times0.1$

$=31.4\Omega$

2. (d) $21.13\Omega$

Explanation:

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

$=\sqrt{\text{12}^2+(\text{31.4}-\text{14})^2}$

$=21.13\Omega$

3. (d) 9.46A

Explanation:

$\text{l}_\text{v}=\frac{\text{E}_\text{v}}{\text{Z}}=\frac{200\text{V}}{21.13}$

$= 9.46\text{A}$

4. (c) 55º4'

Explanation:

$\tan\phi=\frac{\text{X}_\text{L}-\text{X}_\text{C}}{\text{R}}$

$=\frac{3.14-14}{12}=1.45$

$\phi\tan^{-1}(1.45)=55^\circ4'$

5. (c) V_L < V_C