Answer.4. ${Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle – {{\tan }^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)}$

Explanation:-

Parallel RC Circuit Impedance

Since there are only two circuit components, R and C, the total impedance can be expressed by the product-over-sum rule.

${Z = \frac{{\left( {R\angle {0^^\circ }} \right)\left( {{X_C}\angle – {{90}^^\circ }} \right)}}{{R – j{X_C}}}}$

By multiplying the magnitudes, adding the angles in the numerator, and converting the denominator to polar form, we get

$Z = \frac{{\left( {R{X_C}} \right)\angle \left( {{0^^\circ } – {{90}^^\circ }} \right)}}{{\sqrt {{R^2} + {X^2}_C} \angle – {{\tan }^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)}}$

Now, by dividing the magnitude expression in the numerator by that in the denominator, and by subtracting the angle in the denominator from that in the numerator

$Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle \left( { – {{90}^^\circ } + {{\tan }^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)} \right)$

Equivalently, this expression can be written as

${Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle – {{\tan }^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)}$

Answer.2. 44.7 ∠−63.4°Ω

Explanation:-

For the Parallel RC circuit, the total Impedance is

$\begin{array}{l} Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle – {\tan ^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)\\ \\ \\ = \frac{{(100)(50)}}{{\sqrt {{{(100)}^2} + {{(50)}^2}} }}\angle – {\tan ^{ – 1}}\left( {\frac{{100}}{{50}}} \right) \end{array}$

= 44.7 ∠−63.4°Ω

Answer.3. 894∠−26.6°Ω

Explanation:-

From the given figure

Resistance = 1 kΩ

Inductance = 2 kΩ

$\begin{array}{l} Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle – {\tan ^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)\\ \\ \\ = \frac{{(1)(2)}}{{\sqrt {{{(1)}^2} + {{(2)}^2}} }}\angle – {\tan ^{ – 1}}\left( {\frac{1}{2}} \right) \end{array}$

Z = 894∠−26.6°Ω

Answer.4. G∠0°

Explanation:-

The conductance, G, is the reciprocal of resistance. The phasor expression for conductance is expressed as

G = 1/R∠0° = G∠0°

Answer.1. jBC

Explanation:-

Capacitive susceptance (B_C) is the reciprocal of capacitive reactance. The phasor expression for capacitive susceptance is

B_C = 1/X_C∠−90°

B_C = ∠90° = jB_C

Answer.4. Y∠±θ

Explanation:-

Admittance (Y) is the reciprocal of impedance. The phasor expression for admittance is

Y = 1/Z∠ ±θ

Y = ∠ ±θ

Answer.2. $Y = \sqrt {{G^2} + {B^2}_C} $

Explanation:-

In working with parallel circuits, it is often easier to use conductance (G), capacitive susceptance, and admittance (Y) rather than resistance (R), capacitive reactance and impedance (Z).

In a parallel RC circuit, as shown in Figure a, the total admittance is simply the phasor sum of the conductance and the capacitive susceptance.

Y = G + jB_C

$Y = \sqrt {{G^2} + {B^2}_C} $

Answer.3. 3.03 + j1.38

Explanation:-

Resistance R = 330Ω

Conductance = 1/R = 1/330 = 3.03 mS.

Now Capacitive reactance

X_c = 1/2πfC

X_C = 1/2π × 1000 × 0.22 = 723 kΩ

Capacitive Suspectance

B_C = 1/X_C = 1/723 = 1.38 mS.

The total Admittance is

Y_t = G + JB_C

Y_t = 3.03 + j1.38

Answer.3. 300∠−24.5° mS

Explanation:-

Resistance R = 330Ω

Conductance = 1/R = 1/330 = 3.03 mS.

Now Capacitive reactance

X_c = 1/2πfC

X_C = 1/2π × 1000 × 0.22 = 723 kΩ

Capacitive Suspectance

B_C = 1/X_C = 1/723 = 1.38 mS.

The total Admittance is

Y_t = G + JB_C

Y_t = 3.03 + j1.38

Converting to polar form

$\begin{array}{l} {Y_t} = \sqrt {{G^2} + {B^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{B_c}}}{G}} \right)\\ \\ = \sqrt {{{(3.03)}^2} + {{(1.38)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{1.38}}{{3.03}}} \right) \end{array}$

Now total impedance

Z_t = 1/Y_t

Z_t = 1/(3.03 + j1.38)

Z_t = 300∠−24.5° mS

Answer.4. 5∠24.5° mA

Explanation:-

Resistance R = 330Ω

Conductance = 1/R = 1/2.2 = 455 µS.

Now Capacitive reactance

X_c = 1/2πfC

X_C = 1/2π × 1.5 × 0.022 = 4.82 kΩ

Capacitive Suspectance

B_C = 1/X_C = 1/4.82 = 207 µS.

The total Admittance is

Y_t = G + JB_C

Y_t = 455 + j207

Converting to polar form

$\begin{array}{l} {Y_t} = \sqrt {{G^2} + {B^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{B_c}}}{G}} \right)\\ \\ = \sqrt {{{(455)}^2} + {{(207)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{207}}{{455}}} \right) \end{array}$

Y_t = 500∠24.5° mS

Use Ohm’s law to determine the total current.

I_tot = V_s.Y_t

I_tot = (10∠0°)(500∠24.5°)

I_tot = 5∠24.5° mA