Correct answer is 4

To calculate the power dissipation in the circuit, we follow a systematic approach. We're provided with the reactance of the capacitor ($X_C=4\sqrt{3}\mathrm{\Omega }$), the resistance ($R=4\mathrm{\Omega }$), and the peak value of the AC voltage source ($V_{peak}=8\sqrt{2}V$). The power dissipated in an AC circuit is primarily through the resistive component, as inductors and capacitors store and release energy but do not dissipate it as heat.

First, we need to determine the effective impedance of the series circuit, which combines the resistance (R) and the capacitive reactance (X_C) in a series configuration. We calculate the impedance (Z) using the formula:

$Z=\sqrt{R^2+X_C^2}$

Plugging in the given values:

$Z=\sqrt{(4{)}^2+(4\sqrt{3}{)}^2}$

$Z=\sqrt{16+48}=\sqrt{64}=8\mathrm{\Omega }$

Next, we convert the peak voltage to RMS (root mean square) voltage because power calculations in AC circuits are performed using RMS values. The formula to convert peak voltage ($V_{peak}$) to RMS voltage ($V_{RMS}$) is:

$V_{RMS}=\frac{V_{peak}}{\sqrt{2}}$

Plugging in the given peak voltage value:

$V_{RMS}=\frac{8\sqrt{2}}{\sqrt{2}}=8\mathrm{V}$

Now, to find the RMS current ($I_{RMS}$) in the circuit, we use Ohm's law as applied to AC circuits, which is $I_{RMS}=\frac{V_{RMS}}{Z}$:

$I_{RMS}=\frac{8}{8}=1\mathrm{A}$

Finally, the power dissipated in the circuit is calculated using the formula for power in resistive components of an AC circuit, which is $P=I_{RMS}^2\times R$:

$P=(1{)}^2\times 4=4\mathrm{W}$

Therefore, the power dissipation in the circuit is 4 W.

Correct answer is 10

Let's first determine the resistance of the coil when connected to a DC supply. The current drawn by the coil in a DC circuit can be used to calculate its resistance using Ohm's law:

$R=\frac{V}{I}$

Given the DC supply voltage $V_{DC}=20\mathrm{V}$ and the current $I_{DC}=5\mathrm{A}$, the resistance $R$ is:

$R=\frac{20\mathrm{V}}{5\mathrm{A}}=4\mathrm{\Omega }$

Next, let's use the information given for the AC supply. When connected to an AC supply, the coil's impedance $Z$ can be determined using the given current. The total voltage and current in an AC circuit are related to the impedance by the formula:

$Z=\frac{V}{I}$

Given the AC supply voltage $V_{AC}=20\mathrm{V}$ and the current $I_{AC}=4\mathrm{A}$, the impedance $Z$ is:

$Z=\frac{20\mathrm{V}}{4\mathrm{A}}=5\mathrm{\Omega }$

The impedance $Z$ of the coil in an AC circuit is composed of both the resistance $R$ and the inductive reactance $X_L$, related by:

$Z=\sqrt{R^2+X_L^2}$

We already know that $R=4\mathrm{\Omega }$. We can now solve for the inductive reactance $X_L$:

$5=\sqrt{4^2+X_L^2}$

Squaring both sides of the equation:

$25=16+X_L^2$

Solving for $X_L$:

$X_L^2=25-16$

$X_L^2=9$

$X_L=\sqrt{9}$

$X_L=3\mathrm{\Omega }$

The inductive reactance $X_L$ is also related to the inductance $L$ and the angular frequency $\omega$ by the formula:

$X_L=\omega L$

where $\omega =2\pi f$. Given the frequency $f=50\mathrm{Hz}$ and using $\pi =3$, we find:

$\omega =2\times 3\times 50$

$\omega =300\mathrm{rad}/\mathrm{s}$

Now we can solve for the inductance $L$:

$X_L=300L$

$3=300L$

$L=\frac{3}{300}$

$L=0.01\mathrm{H}$

Since $1\mathrm{H}=1000\mathrm{mH}$, the self inductance of the coil is:

$L=0.01\mathrm{H}\times 1000\mathrm{mH}/\mathrm{H}=10\mathrm{mH}$

Therefore, the self inductance of the coil is $10\mathrm{mH}$.

Correct answer is 22

To determine the RMS (Root Mean Square) value of the current in the circuit, we start by analyzing the given emf and the capacitive reactance.

The given alternating emf is:

$\mathrm{E}=110\sqrt{2}\sin 100\mathrm{t}\text{volts}$

Here, the peak voltage (or maximum voltage) ${\mathrm{E}}_{\max }$ is:

${\mathrm{E}}_{\max }=110\sqrt{2}\text{volts}$

Next, the RMS value of the voltage, ${\mathrm{E}}_{\mathrm{rms}}$, is obtained by dividing the peak voltage by $\sqrt{2}$:

${\mathrm{E}}_{\mathrm{rms}}=\frac{{\mathrm{E}}_{\max }}{\sqrt{2}}=\frac{110\sqrt{2}}{\sqrt{2}}=110\text{volts}$

We are given a capacitor with a capacitance $C=2\mu \mathrm{F}=2\times {10}^{-6}\text{F}$ and we need to determine the RMS current. The capacitive reactance ${\mathrm{X}}_{\mathrm{C}}$ is given by:

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

where $\omega$ is the angular frequency. From the given formula for emf, we see that:

$\omega =100\text{rad/s}$

Therefore, the capacitive reactance is:

${\mathrm{X}}_{\mathrm{C}}=\frac{1}{100\times (2\times {10}^{-6})}=\frac{1}{200\times {10}^{-6}}=5000\mathrm{\Omega }$

Now, we can calculate the RMS value of the current ${\mathrm{I}}_{\mathrm{rms}}$ using Ohm's law for AC circuits, which states:

${\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{E}}_{\mathrm{rms}}}{{\mathrm{X}}_{\mathrm{C}}}$

Substituting the known values:

${\mathrm{I}}_{\mathrm{rms}}=\frac{110}{5000}=0.022\text{A}=22\text{mA}$

Hence, the RMS value of the current in the circuit is $22\text{mA}$.

Correct answer is 10

To solve this problem, we need to use the concept of resonance in an LCR (inductor-capacitor-resistor) circuit. At resonance, the inductive reactance and capacitive reactance cancel each other out. The condition for resonance in an LCR circuit is given by:

$\omega L=\frac{1}{\omega C}$

Where:

• $\omega$ is the angular frequency
• $L$ is the inductance
• $C$ is the capacitance

Rewriting for angular frequency:

${\omega }^2=\frac{1}{LC}$

The angular frequency $\omega$ is related to the frequency $f$ by:

$\omega =2\pi f$

Substituting this into the equation for angular frequency gives:

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

Therefore, the frequency $f$ can be found by:

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

Given values:

• Inductance, $L=100\mathrm{mH}=100\times {10}^{-3}\mathrm{H}$
• Capacitance, $C=2.5\mathrm{nF}=2.5\times {10}^{-9}\mathrm{F}$

Plug these values into the frequency equation:

$f=\frac{1}{2\pi \sqrt{(100\times {10}^{-3})(2.5\times {10}^{-9})}}$

First, calculate the product of $L$ and $C$:

$L\cdot C=100\times {10}^{-3}\cdot 2.5\times {10}^{-9}=2.5\times {10}^{-10}$

Now, take the square root of the product:

$\sqrt{2.5\times {10}^{-10}}=\sqrt{2.5}\times {10}^{-5}$

Given that ${\mathrm{a}}^2=10$, we have:

$\mathrm{a}=\sqrt{10}$

Since $\sqrt{2.5}=\frac{\sqrt{10}}{2}$:

$\sqrt{2.5}\times {10}^{-5}=\frac{\sqrt{10}}{2}\times {10}^{-5}$

Now substitute back into the frequency formula:

$f=\frac{1}{2\pi (\frac{\sqrt{10}}{2}\times {10}^{-5})}=\frac{1}{\pi \sqrt{10}\times {10}^{-5}}$

Simplify the equation:

$f=\frac{{10}^5}{\pi \sqrt{10}}$

Given that $\pi \approx 3.14$, we get:

$f\approx \frac{{10}^5}{3.14\times 3.162}$

Simplify further:

$f\approx \frac{{10}^5}{9.93}\approx 10\times {10}^3\mathrm{Hz}$

Thus, the frequency of the AC source is approximately:

$10\times {10}^3\mathrm{Hz}$

Correct answer is 250

$\begin{array}{rl}& R=\frac{100}{5}=20\mathrm{\Omega }\\ & Z=\sqrt{R^2+X_L^2}=40\mathrm{\Omega },I_0=\frac{V_0}{Z}=\frac{200}{40}=5\mathrm{A}\\ & P=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi =\frac{V_0{I}_0}{2}\times \frac{R}{Z}=250\mathrm{W}\end{array}$