Physics M.C.Q (1 Marks)

3. V_a​ > V_b​

Explanation:

For series C - R circuit, the impedance $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_\text{C}}$ where $\text{X}_\text{C}=\frac{\text{i}}{\omega\text{C}}$ and current $\text{I}=\frac{\text{V}}{\text{Z}}$

When the capacitor is filled by mica, the capacitance will be increased. If C increases, X_C​ decreases, so the current will increase and 

hence voltage across resistance increases and voltage across capacitor decreases. thus, V_a​ > V_b​

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

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

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)

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

3. The charge on the plates is in phase with the applied voltage.
4. Power delivered to the capacitor is zero.

Solution:

If the alternating voltage is applied to the capacitor, the plate connected to the positive terminal of the source will be at higher potential and the plate connected to the negative terminal of source will be at lower potential. So the plates capacitor is charged.

If $\text{V}=\text{V}_0\sin\omega\text{t},\text{Q}=\text{C V}_0\sin\omega\text{t}$ or we can say that Q and emf are in pahse.

As $\text{P}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\phi$ and in case of a capacitor, $\phi=\frac{\pi}{2}\text{P} = 0,$ or we can say that power delivered to the capacitor is zero.

⇒ P_av = power delivered = 0.

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

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

4. Only L

Explanation:

In an inductor, current lags behind the input voltage by a phase difference of $\frac{\pi}{2}$.

Current and voltage are in same phase in resistor whereas current leads the voltage by $\frac{\pi}{2}$in a capacitor.

So, the circuit must contain an inductor only.

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

Explanation:

In pure inductive circuit, 

voltage leads by $\frac{\pi}{2}$ 

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

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​)

$=100+\text{j}100$

$=100\sqrt{2}\measuredangle45$

$=141\measuredangle45$

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.

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

Explanation:

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

$\Rightarrow\text{i}=\frac{200\sqrt{2}\frac{1}{\sqrt{2}}}{\sqrt{10000^2+\Big(\frac{1}{100\times10^{-6}}-100\times0\Big)^2}}=\frac{200}{\sqrt{2\times10000^2}}$

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

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.

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)

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

Explanation:

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

when C is doubled, L should be halved so that resonant frequency remains unchanged.

2. 16 V

Explanation:

$\text{V}^2={\text{V}^2_\text{R}}+{\text{V}^2_\text{c}}$

$20^2=12^2+\text{V}^2_\text{c}$

$\text{V}_\text{c}=\sqrt{20^2-12^2}$

$\text{V}_\text{c}=16\text{ V}$