Question

Series Circuits

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Answer 1

^Circuits for problems #1 and #2

Answer 2

1. Use the series circuit to answer questions (a)-(e).
a. What is the total voltage across the bulbs?
\[V _{1} + V _{1} = V_{total} \]
\[V_{total} = 6 V\]
b. What is the total resistance of the circuit?
\[R_{total} = 1 + 1 = 2 \Omega\]
c. What is the current in the circuit?
\[(I) = \frac{ V }{ R }\]
\[(I) = \frac{ 6 }{ 2 } = 3 A\]
d. What is the voltage drop across each light bulb? (Remember that voltage drop is calculated by multiplying current in the circuit by the resistance of a particular resistor: V = IR.)
\[V_{1} = (3)(1) = 3 V\]
Voltage drop of 3V for both light bulbs.
e. Draw the path of the current on the diagram
Positive (longer stick of the batter) goes up, round, back to the short end (negative).

Answer 3

1. Use the series circuit to answer questions (a)-(e).
a. What is the total voltage across the bulbs?
\[V _{1} + V _{1} = V_{total} \]
\[V_{total} = 6 V\]
b. What is the total resistance of the circuit?
\[R_{total} = 1 + 1 = 2 \Omega\]
c. What is the current in the circuit?
\[(I) = \frac{ V }{ R }\]
\[(I) = \frac{ 6 }{ 2 } = 3 A\]
d. What is the voltage drop across each light bulb? (Remember that voltage drop is calculated by multiplying current in the circuit by the resistance of a particular resistor: V = IR.)
\[V_{1} = (3)(1) = 3 V\]
Voltage drop of 3V for both light bulbs.
e. Draw the path of the current on the diagram
Positive (longer stick of the batter) goes up, round, back to the short end (negative).

2. Use the series circuit to answer questions (a)-(e).
a. What is the total voltage across the bulbs?
\[V_{1} + V_{1} + V_{1} = V_{total}\]
\[V_{total} = 6 V\]
b. What is the total resistance of the circuit?
\[R_{total} = 1 + 1 + 1 = 3 \Omega\]
c. What is the current in the circuit?
\[(I) = \frac{ 6 }{ 3 } = 2 A\]
d. What is the voltage drop across each light bulb?
\[V = (2)(1) = 2 V\]
Voltage drop of 2V for each light bulb (since same resistance).
e. Draw the path of the current on the diagram
Positive (longer stick of the battery) goes up, round, back to the short end (negative).

Answer 4

3. What happens to the current in a series circuit as more light bulbs are added? Why?

More light bulbs means more resistance, so the value of the current would decrease due to the inverse proportional relationship to the resistance. The current would remain constant through each resistor (light bulb) though.

4. What happens to the brightness in a series circuit as more light bulbs are added? Why?

Voltage is not constant across each resistor (unlike parallel circuits). Since the power determines the brightness, and Power = Current times Voltage, the voltage drop across each resistor in a series circuit results in a lower power for each light bulb, which means there is a decreased brightness.

Answer 5

^Circuits for problems #5 - #9

Answer 6

5. Use the series circuit to answer questions (a), (b), and (c).
a. What is the total resistance of the circuit?
\[R_{total} = 1 + 2 = 3 \Omega\]
b. What is the current in the circuit?
\[(I) = \frac{ 6 }{ 3 } = 2 A\]
c. What is the voltage drop across each resistor?
\[V = IR \]
\[V_{1} = 2(1) = 2 V\]
\[V_{2} = (2)(2) = 4 V\]

Answer 7

6. Use the series to answer questions (a)-(e)
a. What is the total voltage of the circuit?
\[V = 12 V\]
b. What is the total resistance of the circuit?
\[R_{total} = 2 + 2 = 4 \Omega\]
c. What is the current in the circuit?
\[(I) = \frac{ 12 }{ 4 } = 3 A\]
d. What is the voltage drop across each light bulb?
\[V = RI\]
\[V= 2(3) = 6 V\]
For each 2 ohm resistor
e. Draw the path of the current on the diagram
Out the long end, back through the short end

Answer 8

7. Use the series circuit to answer questions (a), (b), and (c). Consider each resistor equal to all others.
a. What is the resistance of each resistor?
\[3R = \frac{ V }{ I }\]
\[3R = \frac{ 3 }{ 0.5 } \]
\[3R = 6\]
\[R = 2 \Omega\]
b. What is the voltage drop across each resistor?
V = RI
\[V = 0.5(2) = 1 V\]
For each resistor, since all same resistance
c. On the diagram, show the amount of voltage in the circuit before and after each resistor?
3-> 2->1-> 0

Answer 9

8. Use the series to answer questions (a)-(d)
a. What is the total resistance of the circuit?
\[R_{total} = 2 + 3 + 1 = 6 \Omega\]
b. What is the current in the circuit?
\[(I) = \frac{ 9 }{ 6 } = 1.5 A\]
c. What is the voltage drop across each resistor?
\[V = RI\]
\[V_{2} = (1.5)(2) = 3 V\]
\[V_{3} = (1.5)(3) = 4.5 V\]
\[V_{1} = (1.5)(1) = 1.5 V\]
d. What is the sum of the voltage drops across the three resistors? What do you notice about this sum?
\[V_{2} + V_{3} + V_{1} = 9 V\]
This is the total voltage of the battery.

Answer 10

9. Use the diagram to answer questions (a), (b), and (c).
a. How much current would be measured in each circuit if each light bulb has a resistance of 6 ohms?
\[R_{total} = 6 + 6 = 12 \Omega\]
\[V_{total, A} = 6 V\]
\[V_{total, B} = 12 V\]
\[I_{total, A} = \frac{ 6 }{ 12 } = 0.5 A\]
\[I_{total, B} = \frac{ 12 }{ 12 } = 1 A\]
b. How much current would be measured in each circuit if each light bulb has a resistance of 12 ohms?
\[R_{total} = 12 + 12 = 24 \Omega\]
\[V_{total, A} = 6 V\]
\[V_{total, B} = 12 V\]
\[I_{total, A} = \frac{ 6 }{ 24 } = 0.25 A\]
\[I_{total, B} = \frac{ 12 }{ 24 } = 0.5 A\]
c. What happens tot he amount of current in a series circuit as the number of batteries increases?

The current increases as more voltage means a higher current due to the directly proportional relationship.