Alternating Current

The potential difference between point X and Y is given by

V_XY = V_X $-$ V_Y

= V₀ sin($\omega$t) $-$ V₀ sin($\omega$t + ${{2\pi } \over 3}$)

= 2V₀ cos($\omega$t + ${{\pi } \over 3}$) sin ($-$${{\pi } \over 3}$)

= $-$$\sqrt 3 $V₀ cos($\omega$t + ${\pi \over 3}$) = $\sqrt 3 $V₀ sin ($\omega$t $-$ ${\pi \over 6}$).

Similarly, the potential difference between point Y and Z is

V_YZ = V_Y $-$ V_Z

= V₀ sin($\omega$t + ${2\pi \over 3}$) $-$ V₀ sin($\omega$t + ${4\pi \over 3}$)

= 2V₀ cos($\omega$t + $\pi$) sin ($-$${\pi \over 3}$)

= $\sqrt3$V₀ cos($\omega$t) = $\sqrt3$V₀ sin($\omega$t + ${\pi \over 2}$),

and the potential difference between point Z and X is

V_ZX = V_Z $-$ V_X

= V₀ sin($\omega$t + ${4\pi \over 3}$) $-$ V₀ sin($\omega$t)

= $\sqrt3$V₀ cos($\omega$ + ${2\pi \over 3}$) = $\sqrt3$V₀ sin($\omega$t + ${7\pi \over 6}$).

The rms value of the potential V = V₀ sin($\omega$t + $\phi$) is given by V^rms = V₀ / $\sqrt2$. Thus, rms values of the potentials are $V_{XY}^{rms} = V_{YZ}^{rms} = V_{ZX}^{rms} = {V_0}\sqrt {3/2} $. Hence, the reading of the voltmeter is independent of the two terminal i.e., reading is same whether it is connected across X-Y or Y-Z or Z-X.

Given V = V₀ sin $\omega$t

At resonant frequency, current and voltage will be in same phase i.e., $\omega L = {1 \over {\omega C}}$

$\therefore$ Resonant frequency $\omega = {1 \over {\sqrt {LC} }}$ ...... (i)

Here, L = 1 $\mu$H = 10^$-$6 H, C = 10^$-$6 F

$\therefore$ $\omega = {1 \over {\sqrt {{{10}^{ - 12}}} }} = {10^6}$ rad s^$-$1

So, option (a) is incorrect.

Since, ${X_C} = {1 \over {\omega C}},\,{X_L} = \omega L$ ....... (ii)

For large $\omega$, X_C $\sim$ 0, so circuit will behave like a inductive circuit.

So, option (d) is incorrect.

From Eqn. (i)

$\omega = {1 \over {\sqrt {LC} }}$

At resonant frequency or the frequency at which the current will be in phase with the voltage is independent of R.

$\therefore$ option (b) is correct.

From Eqn. (ii), at $\omega$ $\sim$ 0, X_C $\sim$ $\infty$ and X_L = 0

$\therefore$ The current flowing through the circuit becomes nearly zero.

So, option (c) is correct.

The current flowing through the capacitor and its charge are given by

$i = {i_0}\cos \omega t$, .... (1)

$Q = \int {idt = \int {{i_0}\cos \omega t\,dt = {{{i_0}} \over \omega }\sin \omega t + k} } $ $ = {{{i_0}} \over \omega }\sin \omega t$, ...... (2)

where we have used the initial condition, Q = 0 at t = 0, to get the integration constant k = 0. The charge attains the maximum value Q_max at t = $\pi$ / (2$\omega$),

Q_max = i₀/$\omega$ = 1/500 = 2 $\times$ 10^$-$3 C.

The key is switched from B to D at the time t = 7$\pi$/6$\omega$. Substitute t = 7$\pi$/6$\omega$ in equations (1) and (2) to get current and charge

i = i₀ cos(7$\pi$/6) = cos($\pi$ + $\pi$/6) = $-$ cos($\pi$/6)

$ = - \sqrt 3 /2$ A,

Q_i = (i₀/$\omega$) sin(7$\pi$/6) = 2 $\times$ 10^$-$3 sin($\pi$ + $\pi$/6)

= $-$ 1 $\times$ 10^$-$3 C.

Note that negative sign in i indicates the counterclockwise direction of current and negative sign in Q_i indicates the negative charge on the upper plate.

The potential across the capacitor is ${V_C} = {{{Q_i}} \over C} = - {{1 \times {{10}^{ - 3}}} \over {20 \times {{10}^{ - 6}}}}$ = $-$ 50 V (upper plate at lower potential). Immediately after A is connected to D, the potential across the resistor is V_R = 100 V and hence the current through it is i = V_R/R = 100/10 = 10 A (counterclockwise). The potential across the capacitor when it is fully charged is equal to the battery emf. Thus, the charge on the capacitor in fully charged condition is

Q_f = V/C = 50/(20 $\times$ 10^$-$6) = 1 $\times$ 10^$-$3 C.

The sign of Q_f indicates the positive charge on the upper plate. Hence, charge that flows through the battery is

Q = Q_f $-$ Q_i = 1 $\times$ 10^$-$3 $-$ ($-$ 1 $\times$ 10^$-$3) = 2 $\times$ 10^$-$3 C.

Here, $\Omega$ = 100 rad/s, L = 0.5 H, C = 100 $\mu$F, V = 20 V

$\therefore$ ${X_L} = \omega L = 100 \times 0.5 = 50\,\Omega $

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

Impedance across capacitor,

${Z_1} = \sqrt {{R^2} + X_C^2} $

$ = \sqrt {{{(100)}^2} + {{(100)}^2}} $

${Z_1} = 100\sqrt 2 \,\Omega $

$\therefore$ ${I_1} = {{20} \over {100\sqrt 2 }} = {1 \over {5\sqrt 2 }}\,A$

Voltage across 100 $\Omega$

$V = {I_1} \times 100 = {1 \over {5\sqrt 2 }} \times 100 = 10\sqrt 2 \,V$

Impedance across inductance,

${Z_2} = \sqrt {{R^2} + {{({X_L})}^2}} = \sqrt {{{(50)}^2} + {{(50)}^2}} $

${Z_2} = 50\sqrt 2 \,\Omega $

$\therefore$ ${I_2} = {{20} \over {50\sqrt 2 }} = {2 \over {5\sqrt 2 }} = {{\sqrt 2 } \over 5}$

Now, voltage across $50\,\Omega = {{\sqrt 2 } \over 5} \times 50 = 10\sqrt 2 $

${I_1} = {1 \over {5\sqrt 2 }}A$ at 45$^\circ$ leading

${I_2} = {{\sqrt 2 } \over 5}A$ at 45$^\circ$ lagging

$\therefore$ Current through circuit

${I_{net}} = \sqrt {I_1^2 + I_2^2} = \sqrt {{{\left( {{1 \over {5\sqrt 2 }}} \right)}^2} + \left( {{{\sqrt 2 } \over 5}} \right)} = 0.3\,A$

In case A,

The capacitive reactance is

$X_C^A = {1 \over {\omega C}}$

Impedance of the circuit is

${Z_A} = \sqrt {{{(R)}^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} $

$I_R^A = {V \over {\sqrt {{{(R)}^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} }}$ ...... (i)

$V_C^A = {{I_R^A} \over {\omega C}} = {V \over {\sqrt {{{(R\omega C)}^2} + 1} }}$ ...... (ii)

In case B,

The capacitive reactance is

$X_C^B = {1 \over {\omega (4C)}} = {1 \over {4\omega C}}$

Impedance of the circuit is

${Z_B} = \sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}} $

$I_R^B = {V \over {\sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}} }}$ ..... (iii)

$V_C^B = {V \over {\sqrt {{{(4R\omega C)}^2} + 1} }}$ ..... (iv)

From (i) and (iii), we conclude that

$I_R^A < I_R^B$

From (ii) and (iv), we conclude that

$V_C^A > V_C^B$