Series Circuits

Solutions to p19

 

1.†††††††† Three 20 W resistors are connected in series across a 120 V generator. What current flows in the circuit?

Rt = 20 + 20 + 20 = 60 W.

Vt = I Rt

120 = I (60)

I = 2 A.

 

2.†††††††† Ten Christmas lights have equal resistances. When connected to a 120 V outlet, a current of 0.5 A flows through the bulbs. What is the resistance of just one bulb?

††††††††††† Vt = I Rt

††††††††††† 120 = 0.5 Rt

††††††††††† Rt = 240 W.

††††††††††† Each light bulb will have 240/10 lights = 24 W.

 

3.†††††††† A lamp with a resistance of 10 W is connected across a 12 V battery. What resistance must be connected to the lamp to create a current of 0.5 A?

††††††††††† Vt = I Rt

12 = 0.5 Rt

Rt = 24 W.

 

R + Rlamp = 24

††††††††††† R = 24 -10 = 14 W.

 

 

4.†††††††† A 20 W resistor and a 30 W resistor are connected in series and placed across a potential difference of 100 V. Find the voltage drop across each resistor.

 

††††††††††† Vt = I Rt

††††††††††† 100 = I (20 + 30)

††††††††††† I = 2 A

 

††††††††††† V1 = I R1.

††††††††††† = 2(20) = 40 V

 

††††††††††† V2 = I R2.

††††††††††† = 2(30) = 60 V

 

 

5.†††††††† Find the voltage across each resistor, as well as the total voltage.

 

a.

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† ††††††††††† 30 W††††††††††††††† A0.10 A††††††††

 

 

 

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† †††† 50 W

 

V1 = 0.10 (30) = 3 V

V2 = 0.10 (50) = 5 V

Vt = 3 + 5 = 8 V

 

b.

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† ††††††††††† 10 W††††††††††††††† A0.10 A††††††††

 

 

 

††††††††††††††††††††††††††††††††††† ††††† 40W ††††††††† ††††††††20 W

V1 = 0.10 (10) = 1 V

V2 = 0.10 (20) = 2 V

V3 = 0.10 (40) = 4 V

Vt = 7 V

 

 

 

6.†††††††† Find the current as well as the voltage across each resistor.

 

 

a.

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† ††††††††††† 30 W††††††††††††††† A††††††

††††††††††† 80 V

 

 

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† †††† 50 W††††††††††

 

††††††††††† Vt = I Rt

††††††††††† 80 = I (30 + 50)

††††††††††† I = 1 A

††††††††††† V1 = 1 (30) = 30 V

V2 = 1 (50) = 50 V

†††††††††††

††††††††††††††††††††††† †††††††††††

 

b.

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† ††††††††††† 100 W††††††††††††† A††††††

††††††††††† 120 V

 

 

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† †††† 150 W††††††††

†††††††††††††††††††††††

Vt = I Rt

††††††††††† 120 = I (100 + 150)

††††††††††† I = 0.48 A

††††††††††† V1 = 0.48 (100) = 48 V

V2 = 0.48 (150) = 72 V

 

 

 

c.

††††††††††† ††††††† 60V†††††††† ††††††††††††††††††††††††††††††††††† 100 W††††††††††† A

 

 

 

††††††††††††††††††††††††††††††††††† ††††† 80W ††††††††† ††††††††20 W

†††††††††††††††††††††††††††††††††††

 

Vt = I Rt

††††††††††† 60 = I (100 + 20+ 80)

††††††††††† I = 0.30 A

††††††††††† V1 = 0.30 (100) = 30 V

V2 = 0.30 (20) = 6 V

V3 = 0.30 (80) = 24 V

 

 

 


††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† †† V=5

 

7.††††††††

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† ††††††††††† ??? W†††††††††††††† A2.0 A††††††††††

††††††††††† ??? V

 

 

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† ††† 2 W

†††††††††††††††††††††††

In the above diagram, find the total voltage and the missing resistance.

 

V2 = I R2 = 2 (2) = 4 V

 

Vt = 5 + 4 = 9 V

 

R1 = V1/I = 5/2 = 2.5 W.

 

8.†††††††† Three known resistances are connected in series to the terminals of a power source.The potential difference at the terminals of the 3.0 Wresistance is 12 V.

 

 

a.†††††††† What is the potential difference of the power source?

 

I = 12V/3.0W = 4.0 A

Vt = IRt = 4.0 (2 + 3 + 4) = 36 V

b.†††††††† What is the voltage drop across the 4.0 W resistor?

 

V3 = 4.0(4.0) = 16 V

 

c.†††††††† What is the voltage drop across the 2.0 W resistor?

 

8.0 V

 

 

9.†††††††† Use the diagram to your right, where V1 = 12 V; V2 = 5.0 V.

 

a.†††††††† What is the reading on voltmeter V3?

 

V1 = Vt = V2 + V3

††††††††††† 12 = 5.0 + V3

††††††††††††††† V3 = 7.0 V

 

b.†††††††† If the current flowing out of the battery was 125 mA, what would be the value of R2?

 

125 mA = 0.125 A

 

V2 = IR2.

5.0 = 0.125 R2.

R2 = 40 W.

 

10.†††††† Flashback

 

In an electric circuit, the potential difference across the terminals of a resistor was set at different levels and the resulting current intensity was measured.The measurements are recorded in the table below.

 

 

Potential Difference

V (V)

 

Current Intensity

I (A)

 

0

 

0

 

5

 

1.0

 

20

 

4.1

 

35

 

7.1

 

40

 

8.1

 

Draw a graph using the above data and then use the graph to determine the resistance of this resistor.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


5††††††††† 10††††††† 15††††††† 20††††††† 25††††††† 30††††††† 35††††††† 40

††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† Voltage(V)

 

Note although the following seems to involve extra work, it is important to place V on the x axis because of the ay the experiment was done (thatís why V is in the first column; it is the independent variable: what we actually controlled in the experiment)

Slope = (6-2)/(30-10)= I/V = G =0.20 S

R = 1/G = 1/0.20 = 5.0 W.