RC Circuit Question and Answers – 2022

Answer.1. Rectangular, Polar

Explanation:-

Rectangular and Polar Forms

Rectangular and polar are two forms of complex numbers that are used to represent phasor quantities. Each has certain advantages when used in circuit analysis, depending on the particular application. A phasor quantity contains both magnitude and angular position or phase. In this text, italic letters such as V and I are used to represent magnitude only, and boldfaced nonitalic letters such as V and I are used to represent complete phasor quantities.

Answer.2. 10 + j6

Explanation:-

(8 + j5) + (2 + j1)

= (8 + 2) + j(5 + 1)

= 10 + j6

Answer.3. 32 − j4

Explanation:-

(20 – j10) + (12 + j6)

= (20 + 12) + j(-10 + 6)

= 32 + j(-4) = 32 − j4

Answer.1. 84.5 mW

Explanation:-

Given

Resistance R = 2 kΩ

Current I = 6.50 mA

Due to capacitance or inductance, we get reactive power (Q).

The real power (S) depends on only resistance.

S = VI = I²/R

S = (6.5)²/2000

P = 0.845 W

P = 84.5 W

Answer.3. 56Ω − j100Ω

Explanation:-

The impedance is simply the capacitive reactance, and the phase angle is −90° because the capacitance causes the current to lead the voltage by 90°.

Z = R − jX_C

Z = 56Ω − j100Ω

Answer.1. 115 ∠− 60.8°Ω

Explanation:-

The impedance in polar form for series RC circuit is

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(56)}^2} + {{(100)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{100}}{{56}}} \right) \end{array}$

Z = 115 ∠− 60.8°Ω

Answer.3. 3.76∠− 57.8°Ω

Explanation:-

The magnitude of the capacitive reactance is

X_C = 1/2πf_C

X_C = 1/2 × 1000 × 0.01

X_C = 15.9 kΩ

The total impedance in rectangular form is

Z = R − jX_C

Z = 10kΩ − j15.9kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(10)}^2} + {{(15.9)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{15.9}}{{10}}} \right) \end{array}$

Z = 18.8 ∠− 57.8° kΩ

Use Ohm’s law to determine the source voltage

V_s = IZ

= (0.2∠0°)(18.8∠-57.8° kΩ)

= 3.76∠− 57.8° Ω

Answer.2. 1.89∠65.5° mA

Explanation:-

X_C = 1/2πf_C

X_C = 1/6.28 × 1.5 × 0.022

X_C = 4.82 kΩ

The total impedance in rectangular form is

Z = R − jX_C

Z = 2.2 kΩ − j4.82 kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(2.2)}^2} + {{(4.82)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{4.82}}{{2.2}}} \right) \end{array}$

Z = 5.30 ∠− 65.5° kΩ

Use Ohm’s law to determine the source current

I = V/Z

= (10∠0°)/5.30 ∠− 65.5°)

= 1.89∠65.5° mA

Answer.2. Decreased

Explanation:-

The magnitude of the capacitive reactance is

X_C = 1/2πf_C

X_C ∝ 1/f

Since the Capacitive reactance is inversely proportional to the frequency, when the frequency increases, the capacitive reactance decreases.

Answer.1. Directly

Explanation:-

As we know, capacitive reactance varies inversely with frequency.

The impedance of the RC circuit is given as

$Z = \sqrt {{R^2} + {X^2}_C} $

When X_C increases, the entire term under the square root sign increases, and thus the magnitude of the total impedance also increases; and when X_C decreases, the magnitude of the total impedance also decreases. Therefore, in a series RC circuit, Z is inversely dependent on frequency.