Physics help assignment 3

Question

 

Question 1

1.1 State Thevenin’s theorem.

1.2 A 12 V battery is connected in series with a 5 Ω resistor (R1). Two resistors, R2 and RL, with values 10 Ω and 6 Ω respectively, are also included in the circuit.

Use Kirchhoff’s method to calculate the current flowing through the load resistors (RL).

Question 2

Various options are given as possible answers to the following questions. Choose the answer and write the letter (A – D) next to the question numbered (2.1 – 2.6) in the answer book.

2.1 What is the applied voltage for a series RLC circuit when iT =  3mA.   _,

             VL = VvC  = 18 v and  R= 1 Kohm?

_ and _?

A 3 V

B 12.37 V

C 34.98 V

D 48 V

 

2.2 How much current will flow in a 100 Hz series RLC circuit if VS = 20 V, RT = 66 Ω and XT = 47 Ω?

A 1.05 A

B 303 mA

C 247 mA

D 107 mA

 

2.3 What is the current phase angle of a parallel RLC circuit when IL = 15.3 A, IC = 0.43 A and IR = 3.5 A?

A 76.7°

B -4.25°

C 88.8°

D -76.7°

 

 

2.4 What is the range between f1 and f2 of an RLC circuit that resonates at 150 kHz and has a Q of 30?

A 100 kHz to 155 kHz

B 147.5 kHz to 152.5 kHz

C 4500 kHz to 295.5 kHz

D 149.97 Hz to 150 Hz

2.5 Series RLC voltage or impedance totals must be calculated by:

A Subtracting the values

B Graphing the values

C Adding vectors

D Multiplying the values

2.6 When _ XL =  XC  the circuit:

A Draws maximum current

B Applied voltage is zero

C Is at resonance

D Draws minimum current

Answer 1

(1.1)

Thevenin's theorem :

Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent voltage source V_th in series connection with an equivalent resistance R_th.

The equivalent voltabe V_th is the voltage obtained at terminals A-B of the network with terminals A-B open circuited.

Answer 2

(1.2)

Step 1:

Draw the circuit as 12 V battery is connected in series with a  resistor R₁.

Consider R₂ and R_L are parallel to the battery.

Using kirchoffs voltage law :

From loop I₁ :

From loop I₂ :

Multiply equation (2) with 3.

Now solve for using equations (1) and (3).

The current through load resistance R_L is 0.86 A.

Solution :

The current through load resistance R_L is 0.86 A.

Answer 3

(2.1)

Step 1:

The total current in the RLC circuit is .

The resistance in the RLC circuit is

The capacitor voltage .

The inductor voltage .

Formula for the source voltage is .

Where capacitor voltage,

            inductor voltage,

            resistor voltage.

The voltage drop due to resistor can be found using .

Source voltage is

Therefore applied voltage is 3V.

Option (A) is correct choice.

Solution :

Option (A) is correct choice.

Answer 4

(2.2)

Step 1:

The source voltage is .

 

The total resistance in the RLC circuit is .

 

 Total reactance in the RLC circuit is .

The current flowing through RLC circuit is .

 

 

Substitute , and .

 

The current flowing through RLC circuit is 246.83 mA.

 

Option (C) is correct choice.

Solution :

Option (C) is correct choice.

Answer 5

(2.3)

Step 1:

The current through the resistor .

The current through the inductor .

The current through the capacitor .

The current phase angle of a parallel RLC circuit is .

The current phase angle is .

Option (D) is correct choice.

Solution :

Option (D) is correct choice.

Answer 6

(2.4)

Step 1:

The resonating frequency is 150 kHz.

 

The quality factor Q = 30.

 

Bandwidth is given as .

 

Where is bandwidth,

 

             Q is quality factor,

 

             is  resonating frequency.

 

Now observe the option.

 

Only option (B) has the 5 KHz difference.

 

Option (B) is correct choice.

 

Solution :

Option (B) is correct choice.

Answer 7

(2.5)

Step 1:

In the Series RLC circuit, the total impedance can be find as the vector sum of the resistance and reactance.

 

Total impendance is .

 

Magnitude of the impedance is .

So option (C)  correct choice.

Solution :

 

Option (C)  correct choice.

Answer 8

(2.6)

Step 1:

 

In RLC circuit :

 

When , circuit is inductive in nature.

 

When , circuit is capacitive in nature.

 

When , circuit is at resonance.

 

So when , the circuit is at resonance.

Option (C) is correct choice.

Solution :

 

Option (C)  correct choice.