RC Circuit Question and Answers – 2022

Answer.4. Can’t be determined

Explanation:-

The magnitude of the capacitive reactance is

X_C = 1/2πf_C

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \\ \\ Z = \sqrt {{R^2} + \frac{1}{{4{\pi ^2}{f^2}_C}}} \end{array}$

If frequency and resistance are halved, we cannot determined the impedance unless the numerical values are given.

Answer.3. 2121 Ω and ∠ = -19.5°

Explanation:-

The magnitude of the capacitive reactance is

X_C = 1/2πf_C

X_C = 1/6.28 × 15 × 0.015

X_C = 0.707 kΩ

The total impedance in rectangular form is

Z = R − jX_C

Z = 2 kΩ − j2.22 kΩ

Converting it to polar form

$\begin{array}{*{20}{l}} {Z = \sqrt {{R^2} + {X^2}_C} \angle – {{\tan }^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)}\\ \begin{array}{l} \\ \end{array}\\ {Z = \sqrt {{{(2)}^2} + {{(0.707)}^2}} \angle – {{\tan }^{ – 1}}\left( {\frac{{0.707}}{2}} \right)} \end{array}$

Z = 2121Ω ∠− 19.5° kΩ

Answer.1. 180°

Explanation:-

If you multiply the positive real value of  +2 by j, the result is +j2. This multiplication has effectively moved the +2 through an angle to the axis.

Similarly, multiplying +2 by −j rotates it −90° to the  −j  axis. Thus, j is considered a rotational operator.

Mathematically, the j operator has a value of √−1.  If +j2 is multiplied by j, you get

j²2 = (√−1) (√−1)(2) = (−1)(2) =−2

This calculation effectively places the value on the negative real axis. Therefore, multiplying a positive real number by j² converts it to a negative real number, which, in effect, is a rotation of 180° on the complex plane.

Answer.2. Decrease

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.

As the frequency is increased, X_C decreases; so less voltage is dropped across the capacitor. Also, Z decreases as X_C decreases, causing the current to increase. An increase in the current causes more voltage across R.

As the frequency is decreased, increases, X_C  so more voltage is dropped across the capacitor. Also, Z increases as X_C increases, causing the current to decrease. A decrease in the current causes less voltage across R.

As the frequency increases, the voltage across Z remains constant because the source voltage is constant. Also, the voltage across C decreases. The increasing current indicates that Z is decreasing. It does so because of the inverse relationship stated in Ohm’s law i.e Z = V_s/I.  The increasing current also indicates that X_C is decreasing (X_C = V_c/I). The decrease in V_c corresponds to the decrease in X_C.

Since X_C is the factor that introduces the phase angle in a series RC circuit, a change in X_C produces a change in the phase angle. As the frequency is increased, becomes smaller, and thus the phase angle decreases. As the frequency is decreased, X_C becomes larger, and thus the phase angle increases.

Answer.4. Current, Voltage

Explanation:-

When a sinusoidal voltage is applied to a series RC circuit, each resulting voltage drop and the current in the circuit are also sinusoidal and have the same frequency as the applied voltage. The capacitance causes a phase shift between the voltage and current that depends on the relative values of the resistance and the capacitive reactance.

When a circuit is purely capacitive, the phase angle between the applied voltage and the total current is 90° with the current leading the voltage. When there is a combination of both resistance and capacitive reactance in a circuit, the phase angle between the applied voltage and the total current is somewhere between 0° and 90° depending on the relative values of the resistance and the capacitive reactance.

Answer.2. Current, voltage

Explanation:-

• The phase angle is the angle between the total current and source voltage. It may be positive or negative depending on whether the source voltage leads or lags the circuit current.
• When a circuit is purely resistive, the phase angle between the applied (source) voltage and the total current is zero.
• When a circuit is purely capacitive, the phase angle between the applied voltage and the total current is with the current leading the voltage.
• When there is a combination of both resistance and capacitive reactance in a circuit, the phase angle between the applied voltage and the total current is somewhere between 0 and 90° depending on the relative values of the resistance and the capacitive reactance.

Answer.2. Z = 150 ∠-90°

Explanation:-

The total impedance in rectangular form is

Z = R − jX_C

As R which is not given, it’s negligible

Z = − j2.22 kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(150)}^2}} \angle – {\tan ^{ – 1}}\left( {150} \right) \end{array}$

Z = 150 ∠-90°

Answer.4. 4.11 V

Explanation:-

The magnitude of the capacitive reactance is

X_C = 1/2πf_C

X_C = 1/6.28 × 1.2 × 0.02

X_C = 6.67 kΩ

The total impedance in rectangular form is

Z = R − jX_C

Z = 12 kΩ − j6.67 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 =13.72 ∠− 28.81° kΩ

Use Ohm’s law to determine the source voltage

V_s = IZ

= (0.3∠0°)(13.72∠-28.81° kΩ)

= 4.116∠− 28.81° Ω

Answer.2. False

Explanation:-

The product of two positive numbers, as well as two negative numbers, is positive, how can we find the square root of a negative number. Or, can we find a number such that, when it is multiplied by itself gives a negative number. Obviously not. But there arise situations where we have to find the square root of negative numbers.

Solving for x from the equation x² +25 = 0 is such an instance. For enabling these operations, is taken and denoted by I, the first letter of the word imaginary (In fact it is the Greek letter iota). Using this notation, we can write the square root of negative numbers.

An imaginary number that when squared gives a negative result ie i(iota). Imaginary numbers do exist. Despite their name, they are not really imaginary at all.

Answer.2. 195.2 W

Explanation:-

True power P = 150W

Reactive power P_r = 125W

Apparent Power P_A

$\begin{array}{l} {P_A} = \sqrt {{P^2} + {P^2}_R} \\ \\ {P_A} = \sqrt {{{(150)}^2} + {{(125)}^2}} \end{array}$

P_A = 195.2 W