Alternating Current

Given, Inductance, $L=0.1 \mathrm{~H}$

Resistance, $R=110 \Omega$;

Potential difference, $V=110 \mathrm{~V}$

Frequency, $f=350 \mathrm{~Hz}$

Phase difference between maximum voltage and maximum current is given as

$ \begin{aligned} \tan \phi & =\frac{X_L}{R} \\ & =\frac{2 \pi f L}{R}=\frac{2 \times \frac{22}{7} \times 350 \times 01}{110}=2 \\ \Rightarrow \quad \tan \phi & =2 \\ \Rightarrow \quad \phi & =\tan ^{-1}(2) \end{aligned}$

Given, $I = 6 + 10\sin \omega t$

${I_{eft}} = \left[ {{{\int\limits_0^T {{I^2}dt} } \over {\int\limits_0^T {dt} }}} \right] = {\left[ {{1 \over T}\int\limits_0^T {{{(6 + 10\sin \omega t)}^2}dt} } \right]^{1/2}}$

$ = {\left[ {{1 \over T}\int\limits_0^T {(36 + 120\sin \omega t + 100{{\sin }^2}\omega t)dt} } \right]^{1/2}}$

But as, ${1 \over T}\int\limits_0^T {\sin \omega t\,dt = 0} $ and ${1 \over T}\int\limits_0^T {{{\sin }^2}\omega t = {1 \over 2}} $

$ \Rightarrow {I_{eff}} = {\left[ {36 + {1 \over 2} \times 100} \right]^{1/2}} = 9.27$

Thus, ${I_{eff}} = 9.27\,A$

Given, R₁ = 8 $\Omega$

R₂ = 2 $\Omega$

L = 5 mH

C = 50 $\mu$F

and V₀ = 20 cos (2000t)

Impedance, $Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} $

As, here, ${X_L} = \omega L = 2000 \times 5 \times {10^{ - 3}} = 10\Omega $

Similarly, ${X_L} = {1 \over {\omega C}} = {1 \over {2000 \times 50 \times {{10}^{ - 6}}}} = 10\Omega $

$\because$ ${X_L} = {X_C}$, Hence, ${i_{\max }} = {{{V_0}} \over Z} = {{20} \over {(8 + 2)}} = 2A$

Hence, ${i_{rms}} = {{{i_{\max }}} \over {\sqrt 2 }} = {2 \over {\sqrt 2 }} = 1.41 \,A$

and $V = {R_2}\,{i_{rms}} = 1.41 \times 2 = 2.82\,V$

Voltage across secondary coil,

= ${{Power\,across\,load} \over {Current\,pas\sin g\,through\,load}}$

$ \Rightarrow {V_2} = {P \over {{I_L}}} = {{60} \over {0.11}} \Rightarrow {V_2} = 545.45$ V

Voltage across primary coil, V₁ = 220V

$\Rightarrow$ V₂ > V₁

It means that step-up transformer is used.

Given, $V_{\mathrm{rms}}=20 \mathrm{~V}$

Voltage across resistance $=12 \mathrm{~V}$

Let voltage across inductor be $V_L$

As we know that,

$\begin{aligned} V^2 & =V_R^2+V_L^2 \\ \Rightarrow \quad V_L^2 & =V^2-V_R^2=(20)^2-(12)^2 \\ & =400-144=256 \mathrm{~V} \\ V_L & =16 \mathrm{~V} \end{aligned}$

Given, resistance of bulb, $R=280 \Omega$

Supply voltage $V_{\text {rms }}=200 \mathrm{~V}$

As, $\quad V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=$ root mean square speed

$\Rightarrow \quad V_0=\sqrt{2} V_{\mathrm{rms}} $

where, $V_0$ is peak voltage,

As, $\quad V=I R$

$\therefore \quad I_0=\frac{V_0}{R}=\frac{\sqrt{2} \times 200}{280}=\frac{5 \sqrt{2}}{7}=1.010 \mathrm{~A}$

Which is nearly 1A.

Given,

Initial resonance frequency, $f_i=f$

Initial capacitance and inductance be $C_1=C$ and $L_1=L$

Final capacitance and inductance one $C_2=4 C$ and $L_2=L$

Let final frequency be $f^{\prime}$.

Since, $\omega=\frac{1}{\sqrt{L C}}=2 \pi f$

where, $\omega$ and $f$ and angular frequency and frequency.

$\begin{aligned} & \therefore \quad f \propto \frac{1}{\sqrt{L C}} \Rightarrow \frac{f}{f^{\prime}}=\sqrt{\frac{L^{\prime} C^{\prime}}{L C}}=\frac{\sqrt{L 4 C}}{\sqrt{L C}}=\sqrt{4}=2 \\ & \therefore \quad f^{\prime}=\frac{f}{2} \end{aligned}$

As shown in phasor diagram, voltage leads the current.

As in purely inductive circuit, current lags behind the voltage by an angle of 90$^\circ$, here angle is not 90$^\circ$ but it is lesser than that.

So, this type of case arise in R-L-C circuit when R-L-C load with inductive reactance is more than the capacitive reactance as shown below.