Answer.1. −300°

Explanation:-

We subtract 360 from the angle to get its corresponding negative angle.

60° = 60 – 360 = − 300°

Answer.3. Increase

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.2. 8.48 ∠45°

Explanation:-

The magnitude of the phasor represented by 6 + j6 is

$\begin{array}{l} C = \sqrt {{A^2} + {B^2}} \\ \\ C = \sqrt {{8^2} + {6^2}} \end{array}$

C = 8.48

Since the phasor is in the first quadrant, therefore, the angle is

θ = tan⁻¹(±B/A)

θ = tan⁻¹(6/6) = 45°

θ is the angle relative to the positive real axis. The polar form of 8 + j6 is

C∠θ = 8.48∠45°

Answer.4. Resistance and capacitive Reactance

Explanation:-

In an RC circuit, the impedance is determined by both the Resistance and the Capacitive reactance combined.

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

Answer.1. 8.66 + j5

Explanation:-

The real part of the phasor represented by 10∠30° is

A = C.cosφ

= 10 cos 30° = 10(0.866) = 8.66

The imaginary part j of the phasor is

jB = jC sinφ

= j10 sin 30° = j10(0.5) = j5

The rectangular form of 10∠30° is

A + jB = 8.66 + j5

Answer.3. 0° to 90°

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.

As shown in Figure, the resistor voltage V_R the capacitor voltage  V_C and the current (I) are all sine waves with the frequency of the source. Phase shifts are introduced because of the capacitance. The resistor voltage and current lead to the source voltage, and the capacitor voltage lags the source voltage. The phase angle between the current and the capacitor voltage is always 90°.

The amplitudes and the phase relationships of the voltages and current depend on the resistance and capacitive reactance values. 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 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.4. 10∠36.9°

Explanation:-

The magnitude of the phasor represented by 8 + j6 is

$\begin{array}{l} C = \sqrt {{A^2} + {B^2}} \\ \\ C = \sqrt {{8^2} + {6^2}} \end{array}$

C = 10

Since the phasor is in the first quadrant, therefore, the angle is

θ = tan⁻¹(±B/A)

θ = tan⁻¹(6/8) = 36.9°

θ is the angle relative to the positive real axis. The polar form of 8 + j6 is

C∠θ = 10∠36.9°

Answer.4. 4th Quadrant

Explanation:-

A phasor can be visualized as forming a right triangle in the complex plane, as indicated in Figure, for each quadrant location. The horizontal side of the triangle is the real value, A, and the vertical side is the j value, B. The hypotenuse of the triangle is the length of the phasor, C, representing the magnitude, and can be expressed, using the Pythagorean theorem, as

$C = \sqrt {{A^2} + {B^2}} $

So from the above figure, it is clear that the number 24 – j8 is located in the fourth quadrant.

Answer.4. Lead

Explanation:-

In a series RC circuit, the current is the same through both the resistor and the capacitor. Thus, the resistor voltage is in phase with the current, and the capacitor voltage lags the current by 90°. Therefore, there is a phase difference of 90° between the resistor voltage, V_R, and the capacitor voltage V_C, as shown in the waveform diagram.

Answer.4. 141 − j141

Explanation:-

The real part of the phasor represented by 10∠30° is

A = C.cosφ

= 200 cos (−45°) = 200(0.707) = 141

The imaginary part j of the phasor is

jB = jC sinφ

= j200 sin (−45°) = 200(−0.707) = −j141

The rectangular form of 10∠30° is

A + jB = 141 −j141