Physics M.C.Q (1 Marks)

2. increases directly with frequency

1. Zero average current.

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

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

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

1. 50 Volt

4. 4 times

3. 50V

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

1. $2\text{C}$

2. $157\Omega$

3. Admittance

4. Zero

3. LC will have to be decreased

2. $\text{R}$

3. $30\pi$

1. Decrease to one-half of the original value

4. lags behind the voltage by $\frac{\pi}{2}$

1. DC.

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

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

2. inversely proportional to frequency

1. A is true but B is false

3. $20\Omega$

2. increases

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

4. $\text{None of these}$

1. 100V

1. $2.2\text{mA}$

2. the current and P.d across the resistance lags behind the P.d. across the inductance by angle $\frac{\pi}{2}$

4. none of these

2. May be zero.

1. $\text{V}=\text{V}_0\sin\omega\text{t}$

3. R = 15Ω, L = 3.5 H, C = 30μF.

2. 25.6V

3. 1.0012

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

2. $\text{Between }50\Omega \text{ to } 70\Omega$

3. $\frac{\text{E}_\text{v}}{\sqrt{\text{R}^2+\omega^2\text{L}^2}}$

1. 20V, 10V, 10V

1. be 4 times

3. 1

2. 2 A

1. Current.
2. Induced emf in the inductor.

2. 0.7A

1. high impedance

3. This is the similar problem as we discussed above. In this problem, the current increases on increasing the frequency of supply.
4. Hence, the reactance of the circuit must be decreased as increase in frequency.
5. So, one element that must be connected is capacitor.
6. 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.
7. Hence current in the circuit increases.
   Importance point: Resistive, Capacitive Circuit $(RC-$ Circuit$)$
   $V_R = iR,$
   $V_C = iX_C,$
   $V_R = iR$
   
8. Applied voltage: $\text{V}=\sqrt{\text{V}^2_\text{R}+\text{V}^2_\text{C}}$
9. Impedance: $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_\text{C}}=\sqrt{\text{R}^2+\Big(\frac{1}{\omega\text{C}}\Big)^2}$
10. Current: $\text{i}=\text{i}_0\sin(\omega\text{t}+\phi)$
11. Peak current: $\text{i}_0=\frac{\text{V}_0}{\text{Z}}=\frac{\text{V}_0}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}=\frac{\text{V}_0}{\sqrt{\text{R}^2+\frac{1}{4\pi^2\text{v}^2\text{C}^2}}}$
12. Phase difference: $\phi=\tan^{-1}\frac{\text{X}_\text{C}}{\text{R}}=\tan^{-1}\frac{1}{\omega\text{CR}}$
13. Power factor: $\cos\phi=\frac{\text{R}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$
14. Leading quantity: Current.

3. 4.6mA

3. time

1. Here, the power factor $\cos\phi\geq0,\text{P}\geq0$.
2. The driving force can give no energy to the oscillator (P = 0) in some cases.
3. The driving force cannot syphon out (P < 0) the energy out of oscillator.

1. It may be defined as cosine of the angle of lag of lead (i.e., $\cos\phi$).
2. It is also defined as the radio of resistance and impedange $\Big(\text{i.e.,} \frac{\text{R}}{\text{Z}}\Big)$.
3. $\text{The radio}=\frac{\text{True power}}{\text{Apparent power}}=\frac{\text{W}}{\text{VA}}=\frac{\text{kW}}{\text{kVA}}=\cos\phi$

1. As power factor, $\cos\phi=\frac{\text{R}}{\text{Z}}$

2. When $\phi=\frac{\pi}{2}$ (in case of L of C), P = 0.
3. From (a), it is clear that P < 0 is not possible.

1. $5\Omega$

3. Inductive

1. For a given power level, there is a lower current.
2. Lower current implies less power loss.

4. It is easy to reduce the voltage at the receiving end using step-down transformers.

Susceptance is the imaginary part of admittance.
The admittance is the inverse of impedance.
Unit of admittance is ohm. Unit of admittance is $ohm^{-1}$ or siemens.
Unit of susceptance is same as of admittance.

1. $\frac{10}{2\pi}\text{Hz}$

2. $10^4\Omega$

4. $155\Omega$

1. 50 Hz

3. About 310V.

3. C, 9C

1. Inductor and capacitor.

4. Resistor, inductor and capacitor.

4. $10^5\Omega$

2. 25.6V

3. 90°

2. May be 1000W.

4. May be less than 1000W.

1. An inductor and a capacitor.

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

4. (d)

3. 311V

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

4. About 10A.

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

4. Only L

4. In a pure inductive circuit emf will be in phase with the current

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

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
$\text{I}_0=\text{I}_\text{S}\sqrt{2}$
$=(0.5)(1.414)=0.707=\frac{1}{\sqrt{2}}\text{A}$

1. If R is increased current will decrease.

2. $10\sqrt{\text{2}}\text{mA}$

3. A.C. only

2. $5\text{mA}$

1. effective resistance due to capacity

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

2. 16 V

3. I = 0.5 A

3. $13.2\Omega$

1. 2

3. 225.08Hz

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

1. 0 V, 8 A

3. 120 V

2. decrease

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

2. 0.5 A