Alternating Current

Clearly first element is resistance and second element is inductor

$ \begin{aligned} R & =X_L=\frac{200}{5}=40 \Omega \\ \therefore Z & =\sqrt{R_f^2+X_L^2}=40 \sqrt{2} \Omega \\ I_{\mathrm{rms}} & =\frac{V_{\mathrm{rms}}}{Z}=\frac{200 / \sqrt{2}}{40 \sqrt{2}}=\frac{5}{2} \mathrm{~A} \end{aligned} $

To determine the potential differences across the resistor $R$, inductor $L$, and capacitor $C$ in a series AC circuit, we can use Ohm's Law and the given impedances.

Given:

• $R = 24 \textrm{ ohm}$


• Inductive reactance $X_L = 36 \textrm{ ohm}$


• Capacitive reactance $X_C = 24 \textrm{ ohm}$


• Current in the circuit $I = 5 \textrm{ amp}$

First, we find the potential difference across each component using Ohm's Law:

1. Potential difference across $R$:

$ V_R = I \cdot R $

$ V_R = 5 \textrm{ amp} \cdot 24 \textrm{ ohm} $

$ V_R = 120 \textrm{ V} $

1. Potential difference across $L$:

$ V_L = I \cdot X_L $

$ V_L = 5 \textrm{ amp} \cdot 36 \textrm{ ohm} $

$ V_L = 180 \textrm{ V} $

1. Potential difference across $C$:

$ V_C = I \cdot X_C $

$ V_C = 5 \textrm{ amp} \cdot 24 \textrm{ ohm} $

$ V_C = 120 \textrm{ V} $

Summarizing, the potential differences are:

• $V_R = 120 \textrm{ V}$


• $V_L = 180 \textrm{ V}$


• $V_C = 120 \textrm{ V}$

Therefore, the correct answer is:

Option D: $120, 180, 120 \textrm{ V}$

The resulting current is the sum of the direct current and the alternating current:

$I = I_{dc} + I_{ac} = 10 + 10 \sin \omega t$

To find the effective value (also known as the root-mean-square or RMS value), we need to calculate the square root of the mean of the squared current:

$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$

where T is the period of the alternating current.

Let's substitute the expression for I and simplify:

$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} (10 + 10 \sin \omega t)^2 dt}$

Expanding the square and integrating term by term:

$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} (100 + 200 \sin \omega t + 100 \sin^2 \omega t) dt}$

We know that the average value of $ \sin \omega t $ over a complete cycle is zero, and the average value of $ \sin^2 \omega t $ over a complete cycle is $\frac{1}{2}$. Therefore:

$I_{rms} = \sqrt{\frac{1}{T} (100T + 0 + 100 \cdot \frac{1}{2} T)} = \sqrt{150}$

Simplifying further:

$I_{rms} = 5 \sqrt{6} \mathrm{~A}$

Therefore, the correct answer is Option B: $5 \sqrt{6} \mathrm{~A}$.

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

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.

As coil 2 with inductance L₂ has been wounded by a similar wire as coil 1 but the direction of windings is reversed in each layer, so flux through L₂ will be zero.

As, L₂ $\propto$ $\phi$ $\Rightarrow$ L₂ = 0

Also, since the dimensions, number of turns and number of layers of windings are equal in coil 1 and 3.

So, L₁ = L₃

Given, $\omega$ = 2000 rad s^$-$1

L = 5 mH = 5 $\times$ 10^$-$3 H

C = 50 $\mu$F = 50 $\times$ 10^$-$6 F

V = 30 V

Inductive reactance,

X_L = $\omega$L = 2000 $\times$ 5 $\times$ 10^$-$3 = 10 $\Omega$

and capacitive reactance,

${X_C} = {1 \over {\omega C}} = {1 \over {50 \times {{10}^{ - 6}} \times 2000}} = {{100} \over {10}} = 10\Omega $

$\Rightarrow$ X_L = X_C

The given circuit is resonant circuit.

Hence, impedance of circuit, Z = R = 4 + 6 + 0.5 = 10.5$\Omega$

Amplitude of current, ${I_0} = {V \over Z} = {{30} \over {10.5}} = 2.85$ A