Building Circuits - Free Printable

Problem Analysis and Solution

#### Circuit 3: Motor Circuit
The circuit consists of:
- Two cells (batteries)
- A motor
- A 10 Ω resistor
- A switch

Question: What would happen to the speed of the motor if the 10 Ω resistor was replaced with a 20 Ω resistor? Use ideas about resistance, energy, and current in your answer.

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Step-by-Step Solution for Circuit 3

1. Understanding the Circuit:
- The two cells provide a voltage source. Assuming each cell provides 1.5 V, the total voltage across the circuit is:
\[
V_{\text{total}} = 1.5 \, \text{V} + 1.5 \, \text{V} = 3 \, \text{V}
\]
- The motor and the resistor are in series. The total resistance in the circuit is the sum of the motor's internal resistance (\(R_M\)) and the external resistor (\(R\)).

2. Current in the Circuit:
- Using Ohm's Law:
\[
I = \frac{V}{R_{\text{total}}}
\]
where \(R_{\text{total}} = R_M + R\).

- Initially, with a 10 Ω resistor:
\[
R_{\text{total, initial}} = R_M + 10 \, \Omega
\]
\[
I_{\text{initial}} = \frac{3 \, \text{V}}{R_M + 10 \, \Omega}
\]

- If the 10 Ω resistor is replaced with a 20 Ω resistor:
\[
R_{\text{total, new}} = R_M + 20 \, \Omega
\]
\[
I_{\text{new}} = \frac{3 \, \text{V}}{R_M + 20 \, \Omega}
\]

3. Effect on Current:
- Since the total resistance increases when the resistor is changed from 10 Ω to 20 Ω, the current in the circuit decreases:
\[
I_{\text{new}} < I_{\text{initial}}
\]

4. Effect on Motor Speed:
- The speed of a motor is directly proportional to the current flowing through it. This is because the torque produced by the motor, which drives its rotation, depends on the current.
- With a lower current (\(I_{\text{new}}\)), the motor will experience less torque, resulting in a slower speed.

5. Energy Considerations:
- The power delivered to the motor is given by:
\[
P = I^2 \cdot R_M
\]
where \(R_M\) is the internal resistance of the motor.
- Since the current decreases when the resistor is increased, the power delivered to the motor also decreases. This reduced power means less energy is available to drive the motor, further contributing to the decrease in speed.

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Conclusion for Circuit 3:

If the 10 Ω resistor is replaced with a 20 Ω resistor, the total resistance in the circuit increases, leading to a decrease in current. As a result, the motor will experience less torque and will rotate at a slower speed.

Final Answer for Circuit 3:
\[
\boxed{\text{The speed of the motor would decrease.}}
\]

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Circuit 4: Bulb Circuit

The circuit consists of:
- Two cells (batteries)
- A bulb
- A switch
- An ammeter to measure current

Questions:
1. What is the current in the circuit when the switch is open?
2. What is the current in the circuit when the switch is closed?

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Step-by-Step Solution for Circuit 4

1. When the Switch is Open:
- When the switch is open, the circuit is not complete. There is no path for current to flow.
- According to Ohm's Law, current can only flow when there is a closed loop. Since the circuit is open, the current is zero.
- Therefore:
\[
I_{\text{open}} = 0 \, \text{Amps}
\]

2. When the Switch is Closed:
- When the switch is closed, the circuit is complete, and current can flow.
- The total voltage across the circuit is:
\[
V_{\text{total}} = 1.5 \, \text{V} + 1.5 \, \text{V} = 3 \, \text{V}
\]
- Let the resistance of the bulb be \(R_B\). Using Ohm's Law:
\[
I_{\text{closed}} = \frac{V_{\text{total}}}{R_B} = \frac{3 \, \text{V}}{R_B}
\]
- Without the specific value of \(R_B\), we cannot calculate the exact numerical value of the current. However, we know that current will flow when the switch is closed.

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Conclusion for Circuit 4:

1. When the switch is open, the current is:
\[
\boxed{0 \, \text{Amps}}
\]
2. When the switch is closed, the current is:
\[
\boxed{\frac{3 \, \text{V}}{R_B} \, \text{Amps}}
\]
(where \(R_B\) is the resistance of the bulb). If the resistance of the bulb is provided, this expression can be evaluated numerically.