RC Circuits - General Questions

Directions to Solve

RC Circuits - General Questions

1.

When the frequency of the voltage applied to a series RC circuit is decreased, the impedance

A.	increases
B.	decreases
C.	remains the same
D.	doubles

Answer: Option A

Explanation:

No answer description available for this question.

2.

A capacitor with 150 of capacitive reactance is across an ac source. The impedance, expressed in polar form, is

A.	Z = 150
B.	Z = 150∠90°
C.	Z = 150∠ –90°
D.	Z = 150∠180°

Answer: Option C

Explanation:

No answer description available for this question.

3.

A 2 k resistor and a 0.002 F capacitor are in series across an ac source. Current in the circuit is 6.50 mA. The true power is

A.	84.5 mW
B.	845 mW
C.	13 mW
D.	130 mW

Answer: Option A

Explanation:

No answer description available for this question.

4.

A 47 resistor and a capacitor with 150 of capacitive reactance are in series across an ac source. The impedance, expressed in rectangular form, is

A.	Z = 47Ω + j150 Ω
B.	Z = 47Ω – j150 Ω
C.	Z = 197 Ω
D.	Z = 103 Ω

Answer: Option B

Explanation:

No answer description available for this question.

5.

A 120 resistor is in parallel with a capacitor with a capacitive reactance of 40 . Both components are across a 12 V ac source. What is the magnitude of the total impedance?

A.	37.9
B.	3.7
C.	14,400
D.	4,800

Answer: Option A

Explanation:

No answer description available for this question.

6.

The complex number 40 55° is equivalent to

A.	55 + j55
B.	40 + j40
C.	45.88 + j65.52
D.	22.94 + j32.76

Answer: Option D

Explanation:

No answer description available for this question.

7.

A 12 k resistor is in series with a 0.02 F capacitor across a 1.2 kHz ac source. If the current is expressed in polar form as I = 0.30° mA, what is the source voltage expressed in polar form?

A.	45.6 V
B.	411.3 V
C.	0.411 V
D.	4.11 V

Answer: Option D

Explanation:

No answer description available for this question.

8.

A 2 k resistor is in series with a 0.015 F capacitor across a 15 kHz ac source. What is the magnitude of the total impedance and the phase angle?

A.	73.4 Ω and θ = –19.5°
B.	707 Ω and θ = –19.5°
C.	2121 Ω and θ = –19.5°
D.	734 Ω and θ = –38.9°

Answer: Option C

Explanation:

Xc = 1/2πfC = 707 Ω

Z = R - jXc = 2000 - j(707)

|z| = sqrt(R*R + Xc*Xc) = 2121 Ω

Phase angle = arctangent(Xc/R) = 19.5°

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