answers - matthewfrench.net · answers continued 3. a) v 1 v 2 = r 1 r 2 3.1 8.7 = 300 r 2 r 2 =...

2

Answers 1. a) R T = R 1 + R 2 = 10 + 35 = 45Ω b) I = V R = 12 45 = 0.27Ω c) V A→B = ( 10 45 ) V in = 2.7V V B→C = ( 35 45 ) V in = 9.3V V A→C = 2.7 + 9.3 = 12.0V d) V B→C = ( 35 35 + 350 ) 12 = 1.1V 2. a) V 1 V 2 = R 1 R 2 8 16 = R 1 200 R 1 = 100Ω b) R = V I = 100Ω c) When the lamp is added in parallel to R 1 , the combined resistance of the components in parallel is: 1 R T = 1 100 + 1 100 , R T = 50Ω. These components will both receive the same share of V in according to the resistance value of 50Ω. The lamp now receives V out = 50 250 x 24 = 4.8V due to the decrease in the combined resistance of R 1 and the lamp. Therefore, the lamp does not operate at its correct brightness. Questions - Potential Divider Circuits

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Page 1: Answers - matthewfrench.net · Answers Continued 3. a) V 1 V 2 = R 1 R 2 3.1 8.7 = 300 R 2 R 2 = 841Ω b) 15ºC c) I = V R = 12 841+300 = 13.8mA d) As the temperature increases, the

Answers

1. a) RT = R1 + R2 = 10 + 35 = 45Ω

b) I =VR

=1245

= 0.27Ω

c) VA→B = ( 1045 ) Vin = 2.7V

VB→C = ( 3545 ) Vin = 9.3V

VA→C = 2.7 + 9.3 = 12.0V

d) VB→C = ( 3535 + 350 ) 12 = 1.1V

2. a) V1

V2=

R1

R2

816

=R1

200

R1 = 100Ω

b) R =VI

= 100Ω

c) When the lamp is added in parallel to R1, the combined resistance of the components in parallel is: 1RT

=1

100+

1100

, RT = 50Ω. These components will both receive the same share of Vin according to the

resistance value of 50Ω.

The lamp now receives Vout =50250

x 24 = 4.8V due to the decrease in the combined resistance of R1 and the

lamp. Therefore, the lamp does not operate at its correct brightness.

Questions - Potential Divider Circuits

Page 2: Answers - matthewfrench.net · Answers Continued 3. a) V 1 V 2 = R 1 R 2 3.1 8.7 = 300 R 2 R 2 = 841Ω b) 15ºC c) I = V R = 12 841+300 = 13.8mA d) As the temperature increases, the

Answers Continued

3. a) V1

V2=

R1

R2

3.18.7

=300R2

R2 = 841Ω

b) 15ºC

c) I =VR

=12

841 + 300= 13.8m A

d) As the temperature increases, the resistance of the thermistor decreases.

The share of the voltage across the thermistor will decrease, thus the reading on the voltmeter increases.

The total resistance of the circuit will decrease, thus the reading on the ammeter will increase.

e) This circuit could be used to automatically turn on a fan if the temperature gets too large.

Questions - Potential Divider Circuits

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