Physics M.C.Q (1 Marks)

2. May be 1000W.

4. May be less than 1000W.

Explanation:

Average power $\text{P}_\text{av}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\phi$

$=100\times10\cos\phi$

$\text{P}_\text{av}=1000\cos\phi$

$\therefore\ \cos\phi$ lies "0 to 1".

$\Rightarrow\ 0\leq\text{P}_\text{av}\leq1000.$

2. May be zero.

Explanation:

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

$\omega=2\pi\text{f}=2\times3.14\times50$

$\omega=314$

$\text{V}_\text{avg}=\frac{\int\limits_0^{0.01}\text{V}\text{dt}}{\int\limits_0^{0.01}\text{dt}}$

$=\text{V}_0\Big(\frac{1\cos\omega\text{t}}{\omega}\Big)_0^{0.01}$

$=\frac{\text{V}_0}{\omega\times0.01}\big(1-\cos\omega(0.1)\big)$

$=\frac{\text{V}_0}{314\times0.01}\big(1-\cos(314\times0.01)\big)$

$=\frac{\text{V}_0}{3.14}\big(1-\cos(314)\big)$

$=\frac{\text{V}_0}{3.14}\big(1-\cos\pi\big)$

$=\frac{2\text{V}_0}{\pi}=140.127\text{volt}$

if $\text{V}=\text{V}_0\cos\omega\text{t}$

$\text{V}_\text{avg}=\frac{\int\text{V d}\rho}{\int\text{dt}}=0$

3. i₁ < i₂

Explanation:

$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$\text{i}=\frac{\text{dQ}}{\text{dt}}$

$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$\in_0\text{C}\times100\pi\Big[\sin\big(100\pi\text{s}^{-1}\big)\text{t}\Big]\\+\in_0\text{C}\times500\pi\Big[\sin\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$

$=100\text{C}\pi\in_0\cos\Big[\big(100\pi\text{s}^{-1}\big)\text{t}+\phi_1\Big]\\+500\text{C}\pi\in_0\cos\Big[\big(500\pi\text{s}^{-1}\big)\text{t}+\phi_2\Big]$

$\text{i}_1=100\pi\in_0\text{C}$ and $\text{i}_2=500\pi\in_0\text{C}$

$\text{i}_2>\text{i}_1$

1. $5\Omega$

Explanation:

$\text{Z}=\sqrt{\text{R}^2+\text{X}^2}$

$\text{R}=4\Omega,\text{X}=3\Omega$

$=\text{Z}=\sqrt{4^2+3^2}=5\Omega$

3. About 310V.

Explanation:

$\text{V}_\text{rms}=220\text{V}$

$\text{V}_\text{p}=\sqrt{2}\times\text{V}_\text{rms}$

$=220\times1.414=311\text{volt}$

1. An inductor and a capacitor.

Explanation:

$\text{X}=0$ (Given)

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

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

It is possible that the circuit contains an inductor and a capacitor.

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.

1. Of the inductor increases.

Explanation:

$\text{X}_\text{L}=\omega\text{L}$

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

If frequency increases that causes 'X_L' raction of inductor increases and 'X_C' reactance of capacitor decreses.

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

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

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