Alternating Current

$i\left(t\right)=i_0\sin 100\pi t$

$\frac{i_0}{2}=i_0\sin 100\pi t_1$

$\sin 100\pi t_1=\sin \frac{\pi }{6}$

$100\pi t_1=\frac{\pi }{6}$

$t_1=\frac{1}{600}\text{sec}$

Similarly $t_2=\frac{1}{200}\text{sec}$

$t=t_2-t_1$

$=\frac{1}{200}-\frac{1}{600}$

$t=\frac{3-1}{600}=\frac{1}{300}$

t = 3.3 ms

The power output of the transformer is the power consumed by the bulb, which is 100 W. The efficiency of the transformer is given as 0.9.

First, we calculate the power input to the transformer using the efficiency formula:

\(\eta = \frac{P_{out}}{P_{in}}\)

where \(\eta\) is the efficiency, \(P_{out}\) is the output power, and \(P_{in}\) is the input power.

Rearranging for \(P_{in}\):

\(P_{in} = \frac{P_{out}}{\eta}\)

Substituting the given values:

\(P_{in} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\)

Next, we use the input power and input voltage to find the input current. The input voltage is given as 220 V.

Using the power formula:

\(P_{in} = V_{in} \times I_{in}\)

Rearranging for \(I_{in}\):

\(I_{in} = \frac{P_{in}}{V_{in}}\)

Substituting the values:

\(I_{in} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\)

Thus, the input current drawn from the mains is approximately \(\frac{5}{9} \text{ A}\).
Identify the power consumed by the bulb: \(P_{out} = 100 \text{ W}\).

Calculate the input power using the efficiency: \(P_{in} = \frac{P_{out}}{\eta} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\).

Identify the input voltage: \(V_{in}\) = \(220 \text{ V}\).

Use the power formula to find the input current: \(P_{in} = V_{in} \times I_{in}\).

Rearrange and solve for \(I_{in}: I_{in} = \frac{P_{in}}{V_{in}} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\).

final answer: \(\frac{5}{9} \text{ A}\)

Given the current as $I=I_A\sin \omega t+$ $I_B\cos \omega t$, the r.m.s. current is calculated by integrating the square of the current over time, then taking the square root. i.e.,

$$\begin{gathered} i_{ms}=\sqrt{\frac{\int I^{2}dt}{\int dt}} \\ I_{rms}=\sqrt{\frac{I_A^2+I_B^2}{2}} \end{gathered}$$

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

, thus:

To find the average power over a cycle, integrate this expression over one period

, then divide by

:

This separates into two integrals:

\begin{equation}
\begin{aligned}
& X_L=45 \mathrm{~W}, X_C=25 \mathrm{~W}, R=20 \mathrm{~W} \\
& I=\frac{220}{\sqrt{\left(X_L-X_C\right)^2+R^2}}=\frac{220}{\sqrt{(45-25)^2+20^2}} \\
& =\frac{220}{2 \sqrt{2}}=\frac{11}{\sqrt{2}}=7.779 \mathrm{~A} \\
& \tan \phi=\frac{X_L-X_C}{R}=\frac{45-25}{20}=1 \\
& f=45^{\circ}
\end{aligned}
\end{equation}

In an ideal transformer, the quantity that remains the same in the primary and secondary coils is power.
Explanation:
Power conservation: An ideal transformer operates on the principle of conservation of energy, meaning the power input to the primary coil is equal to the power output from the secondary coil.

\begin{aligned}
& \text{Given: Reactance of a capacitor } X = \frac{1}{\omega C} \\[4pt]
& \text{If capacitance is doubled and frequency is tripled:} \\[2pt]
& C' = 2C,\quad \omega' = 3\omega \\[4pt]
& \text{New reactance: } X' = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6\,\omega C} \\[4pt]
& \text{Since } X = \frac{1}{\omega C},\; \Rightarrow X' = \frac{X}{6} \\[6pt]
& \boxed{X' = \dfrac{X}{6}}
\end{aligned}

\begin{aligned}
& \text{Given: Reactance of a capacitor } X = \frac{1}{\omega C} \\[4pt]
& \text{If capacitance is doubled and frequency is tripled:} \\[2pt]
& C' = 2C,\quad \omega' = 3\omega \\[4pt]
& \text{New reactance: } X' = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6\,\omega C} \\[4pt]
& \text{Since } X = \frac{1}{\omega C},\; \Rightarrow X' = \frac{X}{6} \\[6pt]
& \boxed{X' = \dfrac{X}{6}}
\end{aligned}

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

This separates into two integrals:

$V_{rms}=\frac{V_0}{\sqrt{2}}=\frac{20}{\sqrt{2}}, I_{rms}=\frac{I_0}{\sqrt{2}}=\frac{10}{\sqrt{2}}$

$<P>=\frac{20}{\sqrt{2}}\times \frac{10}{\sqrt{2}}\times \cos \frac{\mathrm{\pi }}{3}$

$=\frac{200}{2}\times \frac{1}{2}=50W$

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

, thus:

To find the average power over a cycle, integrate this expression over one period

, then divide by

:

This separates into two integrals:

In an ideal transformer, the quantity that remains the same in the primary and secondary coils is power.
Explanation:
Power conservation: An ideal transformer operates on the principle of conservation of energy, meaning the power input to the primary coil is equal to the power output from the secondary coil.

\begin{aligned}
& \text{Given: Reactance of a capacitor } X = \frac{1}{\omega C} \\[4pt]
& \text{If capacitance is doubled and frequency is tripled:} \\[2pt]
& C' = 2C,\quad \omega' = 3\omega \\[4pt]
& \text{New reactance: } X' = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6\,\omega C} \\[4pt]
& \text{Since } X = \frac{1}{\omega C},\; \Rightarrow X' = \frac{X}{6} \\[6pt]
& \boxed{X' = \dfrac{X}{6}}
\end{aligned}

The power output of the transformer is the power consumed by the bulb, which is 100 W. The efficiency of the transformer is given as 0.9.

First, we calculate the power input to the transformer using the efficiency formula:

\(\eta = \frac{P_{out}}{P_{in}}\)

where \(\eta\) is the efficiency, \(P_{out}\) is the output power, and \(P_{in}\) is the input power.

Rearranging for \(P_{in}\):

\(P_{in} = \frac{P_{out}}{\eta}\)

Substituting the given values:

\(P_{in} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\)

Next, we use the input power and input voltage to find the input current. The input voltage is given as 220 V.

Using the power formula:

\(P_{in} = V_{in} \times I_{in}\)

Rearranging for \(I_{in}\):

\(I_{in} = \frac{P_{in}}{V_{in}}\)

Substituting the values:

\(I_{in} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\)

Thus, the input current drawn from the mains is approximately \(\frac{5}{9} \text{ A}\).
Identify the power consumed by the bulb: \(P_{out} = 100 \text{ W}\).

Calculate the input power using the efficiency: \(P_{in} = \frac{P_{out}}{\eta} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\).

Identify the input voltage: \(V_{in}\) = \(220 \text{ V}\).

Use the power formula to find the input current: \(P_{in} = V_{in} \times I_{in}\).

Rearrange and solve for \(I_{in}: I_{in} = \frac{P_{in}}{V_{in}} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\).

final answer: \(\frac{5}{9} \text{ A}\)

\begin{aligned}
& \text{Given: Reactance of a capacitor } X = \frac{1}{\omega C} \\[4pt]
& \text{If capacitance is doubled and frequency is tripled:} \\[2pt]
& C' = 2C,\quad \omega' = 3\omega \\[4pt]
& \text{New reactance: } X' = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6\,\omega C} \\[4pt]
& \text{Since } X = \frac{1}{\omega C},\; \Rightarrow X' = \frac{X}{6} \\[6pt]
& \boxed{X' = \dfrac{X}{6}}
\end{aligned}

\begin{aligned}
& \text{Given: Reactance of a capacitor } X = \frac{1}{\omega C} \\[4pt]
& \text{If capacitance is doubled and frequency is tripled:} \\[2pt]
& C' = 2C,\quad \omega' = 3\omega \\[4pt]
& \text{New reactance: } X' = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6\,\omega C} \\[4pt]
& \text{Since } X = \frac{1}{\omega C},\; \Rightarrow X' = \frac{X}{6} \\[6pt]
& \boxed{X' = \dfrac{X}{6}}
\end{aligned}

The current has maximum amplitude in an LCR circuit at resonance frequency ($\omega_0$), where:

$$\omega_0 = \frac{1}{\sqrt{L \cdot C}}$$​

Given:

$$L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H}, \, C = 10 \, \text{nF} = 10 \times 10^{-9} \, \text{F}$$

Substitute the values into the formula:

$$\omega_0 = \frac{1}{\sqrt{(10 \times 10^{-3})(10 \times 10^{-9})}}$$​

Simplify:

$$\omega_0 = \frac{1}{\sqrt{10 \times 10^{-12}}} = \frac{1}{\sqrt{10^{-11}}} = 10^5 \, \text{rad/s}$$

Hence, the angular frequency of the source is:
$10^5 \, \text{rad/s}$

Correct option: (c).

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

This separates into two integrals:

Current has maximu amplitude at the condition of resonance. In resonance the impedence of the circuit is equal to resistance of the circuit. When resistance is halved, then the maximum amplite of current is doubled.

\begin{aligned}
& \text{Given: Reactance of a capacitor } X = \frac{1}{\omega C} \\[4pt]
& \text{If capacitance is doubled and frequency is tripled:} \\[2pt]
& C' = 2C,\quad \omega' = 3\omega \\[4pt]
& \text{New reactance: } X' = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6\,\omega C} \\[4pt]
& \text{Since } X = \frac{1}{\omega C},\; \Rightarrow X' = \frac{X}{6} \\[6pt]
& \boxed{X' = \dfrac{X}{6}}
\end{aligned}

The initial current amplitude is $I_0$ :

$I_0=\frac{E}{R}$

If R becomes twice its initial value, then:

$R^{\text{'}}=2R$

The new current amplitude $I^{\text{'}}$ will be:

$I^{\text{'}}=\frac{E}{R^{\text{'}}}=\frac{E}{2R}=\frac{I_0}{2}$

Statement I: True. In series LCR, at resonance:

$$X_L=X_C,\quad Z=R$$

So current is maximum:

$I_{\max }=\frac{V}{R}$

Statement II: True. A purely resistive circuit has current $\frac{V}{R}$​, while series LCR has

$$I=\frac{V}{\sqrt{R^2+(X_L-X_C)^2}}$$

So LCR current is never greater than purely resistive current for the same $R$.

The resulatant peak value is

${\mathrm{I}}_{\mathrm{o}} = \sqrt{{\mathrm{I}}_{\mathrm{A}}^2 + {\mathrm{I}}_{\mathrm{B}}^2}$

So, the rms value of current is

${\mathrm{I}}_{\mathrm{rms}} = \frac{{\mathrm{I}}_{\mathrm{o}}}{\sqrt{2}}=\sqrt{\frac{{\mathrm{I}}_{\mathrm{A}}^2 + {\mathrm{I}}_{\mathrm{B}}^2}{2}}$

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

, thus:

To find the average power over a cycle, integrate this expression over one period

, then divide by

:

This separates into two integrals:

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

, thus:

To find the average power over a cycle, integrate this expression over one period

, then divide by

:

This separates into two integrals:

X is a capacitor and its capacitive reactance is equal to the resistance ($X_c=R$Sv6Kpe[]). This is because the current ($I=I_0\sin (\omega t+\frac{\pi }{4})$ is leading the voltage ($v=v_0\sin \left(\omega t\right)$) by a phase angle of $\frac{\pi }{4}$, which is characteristic of a capacitive circuit where current leads voltage. The phase angle ($\varphi$Sv6Kpe[]) in an R-X series circuit is given by

The resulatant peak value is

${\mathrm{I}}_{\mathrm{o}} = \sqrt{{\mathrm{I}}_{\mathrm{A}}^2 + {\mathrm{I}}_{\mathrm{B}}^2}$

So, the rms value of current is

${\mathrm{I}}_{\mathrm{rms}} = \frac{{\mathrm{I}}_{\mathrm{o}}}{\sqrt{2}}=\sqrt{\frac{{\mathrm{I}}_{\mathrm{A}}^2 + {\mathrm{I}}_{\mathrm{B}}^2}{2}}$

Given the current as $I=I_A\sin \omega t+$ $I_B\cos \omega t$, the r.m.s. current is calculated by integrating the square of the current over time, then taking the square root. i.e.,

$$\begin{gathered} i_{ms}=\sqrt{\frac{\int I^{2}dt}{\int dt}} \\ I_{rms}=\sqrt{\frac{I_A^2+I_B^2}{2}} \end{gathered}$$

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

This separates into two integrals:

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

This separates into two integrals:

$i\left(t\right)=i_0\sin 100\pi t$

$\frac{i_0}{2}=i_0\sin 100\pi t_1$

$\sin 100\pi t_1=\sin \frac{\pi }{6}$

$100\pi t_1=\frac{\pi }{6}$

$t_1=\frac{1}{600}\text{sec}$

Similarly $t_2=\frac{1}{200}\text{sec}$

$t=t_2-t_1$

$=\frac{1}{200}-\frac{1}{600}$

$t=\frac{3-1}{600}=\frac{1}{300}$

t = 3.3 ms

In an ideal transformer, the quantity that remains the same in the primary and secondary coils is power.
Explanation:
Power conservation: An ideal transformer operates on the principle of conservation of energy, meaning the power input to the primary coil is equal to the power output from the secondary coil.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{p}} = 220 \mathrm{V}, {\mathrm{N}}_{\mathrm{p}} = 500, {\mathrm{N}}_{\mathrm{s}} = 5000 \\ \frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}}=\frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}} \\ \frac{{\mathrm{V}}_{\mathrm{s}}}{220}=\frac{5000}{500} \\ \mathrm{Vs} =2200 \mathrm{V} \end{gathered}$$

Frequency remains same.

The power output of the transformer is the power consumed by the bulb, which is 100 W. The efficiency of the transformer is given as 0.9.

First, we calculate the power input to the transformer using the efficiency formula:

\(\eta = \frac{P_{out}}{P_{in}}\)

where \(\eta\) is the efficiency, \(P_{out}\) is the output power, and \(P_{in}\) is the input power.

Rearranging for \(P_{in}\):

\(P_{in} = \frac{P_{out}}{\eta}\)

Substituting the given values:

\(P_{in} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\)

Next, we use the input power and input voltage to find the input current. The input voltage is given as 220 V.

Using the power formula:

\(P_{in} = V_{in} \times I_{in}\)

Rearranging for \(I_{in}\):

\(I_{in} = \frac{P_{in}}{V_{in}}\)

Substituting the values:

\(I_{in} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\)

Thus, the input current drawn from the mains is approximately \(\frac{5}{9} \text{ A}\).
Identify the power consumed by the bulb: \(P_{out} = 100 \text{ W}\).

Calculate the input power using the efficiency: \(P_{in} = \frac{P_{out}}{\eta} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\).

Identify the input voltage: \(V_{in}\) = \(220 \text{ V}\).

Use the power formula to find the input current: \(P_{in} = V_{in} \times I_{in}\).

Rearrange and solve for \(I_{in}: I_{in} = \frac{P_{in}}{V_{in}} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\).

final answer: \(\frac{5}{9} \text{ A}\)

Given, $Q=Q_{0}e{-}^{Rt/2L}$

$R=1.5\Omega , L=12mH, \ln \left(2\right)=0.693$

$Q=0.5, Q_0, t=?$

$\Rightarrow \frac{1}{2}=e^{-Rt/2L}$

Taking log on both sides

$\ln \left(\frac{1}{2}\right)=\ln e^{-Rt/2L}$

$t=\frac{2L\ln 2}{R}=\frac{2\times 12\times {10}^{-3}\times 0.693}{1.5}$

$t=11.09 \mathrm{ms}$