Solved WORKSHEET B: COMBINATION CIRCUITS 1) Rank in order, | Chegg.com - Free Printable

Worksheet B: Combination Circuits

Let's solve each problem step by step.

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#### Problem 1:
Rank in order, from largest to smallest, the three currents \( I_1 \), \( I_2 \), and \( I_3 \).

Circuit Diagram:
- Resistors: \( R_1 = 3R \), \( R_2 = R \), \( R_3 = 2R \)
- The circuit is a combination of series and parallel resistors.

Solution:
1. Identify the structure:
- \( R_1 \) and \( R_2 \) are in series.
- The combination of \( R_1 \) and \( R_2 \) is in parallel with \( R_3 \).

2. Calculate equivalent resistance:
- Series combination of \( R_1 \) and \( R_2 \):
\[
R_{\text{series}} = R_1 + R_2 = 3R + R = 4R
\]
- Parallel combination of \( R_{\text{series}} \) and \( R_3 \):
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3} = \frac{1}{4R} + \frac{1}{2R} = \frac{1 + 2}{4R} = \frac{3}{4R}
\]
\[
R_{\text{parallel}} = \frac{4R}{3}
\]

3. Determine the currents:
- Total current \( I \) (from the voltage source):
\[
I = \frac{V}{R_{\text{parallel}}} = \frac{V}{\frac{4R}{3}} = \frac{3V}{4R}
\]
- Current through \( R_3 \) (\( I_3 \)):
\[
I_3 = \frac{V}{R_3} = \frac{V}{2R}
\]
- Current through \( R_1 \) and \( R_2 \) (\( I_1 \) and \( I_2 \)):
Since \( R_1 \) and \( R_2 \) are in series, the current through them is the same:
\[
I_1 = I_2 = \frac{V}{R_{\text{series}}} = \frac{V}{4R}
\]

4. Compare the currents:
- \( I_3 = \frac{V}{2R} \)
- \( I_1 = I_2 = \frac{V}{4R} \)
- Clearly, \( I_3 > I_1 = I_2 \).

Order:
\[
I_3 > I_1 = I_2
\]

Explanation:
The current through \( R_3 \) is larger because it has a lower resistance compared to the series combination of \( R_1 \) and \( R_2 \).

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#### Problem 2:
Bulbs A, B, and C are identical.

(a) Rank in order, from most to least, the brightness of the three bulbs.

Circuit Diagram:
- Bulbs A, B, and C are connected in a specific configuration.

Solution:
1. Identify the structure:
- Bulb A is in series with the parallel combination of bulbs B and C.
- Bulbs B and C are in parallel.

2. Brightness depends on power:
- Power dissipated by a bulb is given by \( P = I^2 R \).
- Since the bulbs are identical, their resistances are the same (\( R \)).

3. Current distribution:
- Let the total current from the source be \( I \).
- Current through bulb A (\( I_A \)) is the same as the total current \( I \) because it is in series with the rest of the circuit.
- Current through bulbs B and C (\( I_B \) and \( I_C \)) splits equally because they are in parallel:
\[
I_B = I_C = \frac{I_A}{2}
\]

4. Power dissipation:
- Power in bulb A:
\[
P_A = I_A^2 R
\]
- Power in bulbs B and C:
\[
P_B = P_C = I_B^2 R = \left(\frac{I_A}{2}\right)^2 R = \frac{I_A^2 R}{4}
\]

5. Compare the powers:
- \( P_A = I_A^2 R \)
- \( P_B = P_C = \frac{I_A^2 R}{4} \)
- Clearly, \( P_A > P_B = P_C \).

Order:
\[
A > B = C
\]

Explanation:
Bulb A is brighter because it carries the full current from the source, while bulbs B and C share the current and thus have less power dissipation.

(b) Suppose a wire is connected between points 1 and 2. What happens to each bulb? Does it get brighter, stay the same, or go out? Explain.

Solution:
1. Effect of short-circuiting points 1 and 2:
- Connecting points 1 and 2 with a wire creates a short circuit across bulbs B and C.
- This means the resistance of the path through B and C becomes effectively zero.

2. Current redistribution:
- All the current from the source will now flow through the short circuit (points 1 and 2), bypassing bulbs B and C.
- Bulb A will still carry the full current from the source.

3. Brightness changes:
- Bulb A: Remains lit because it is still in the main current path.
- Bulbs B and C: Go out because no current flows through them due to the short circuit.

Summary:
- Bulb A: Stays the same (still lit).
- Bulbs B and C: Go out.

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#### Problem 3:
Currents through parallel circuit:

(a) Rank in order, from largest to smallest, the three currents \( I_1 \), \( I_2 \), and \( I_3 \).

Circuit Diagram:
- Resistors: \( 1\Omega \), \( 2\Omega \), \( 3\Omega \)
- Voltage source: \( 12V \)

Solution:
1. Identify the structure:
- All resistors are in parallel.

2. Current through each resistor:
- Current through \( 1\Omega \) resistor (\( I_1 \)):
\[
I_1 = \frac{V}{R_1} = \frac{12V}{1\Omega} = 12A
\]
- Current through \( 2\Omega \) resistor (\( I_2 \)):
\[
I_2 = \frac{V}{R_2} = \frac{12V}{2\Omega} = 6A
\]
- Current through \( 3\Omega \) resistor (\( I_3 \)):
\[
I_3 = \frac{V}{R_3} = \frac{12V}{3\Omega} = 4A
\]

3. Compare the currents:
- \( I_1 = 12A \)
- \( I_2 = 6A \)
- \( I_3 = 4A \)
- Clearly, \( I_1 > I_2 > I_3 \).

Order:
\[
I_1 > I_2 > I_3
\]

Explanation:
The current is inversely proportional to the resistance. The smaller the resistance, the larger the current.

(b) Calculate the equivalent resistance for the above circuit and the total current. What is the current through the 2-Ohm resistor?

Solution:
1. Equivalent resistance:
- For parallel resistors:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{1\Omega} + \frac{1}{2\Omega} + \frac{1}{3\Omega} = 1 + 0.5 + 0.333 = 1.833
\]
\[
R_{\text{eq}} = \frac{1}{1.833} \approx 0.545\Omega
\]

2. Total current:
- Total current from the source:
\[
I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{12V}{0.545\Omega} \approx 22.02A
\]

3. Current through the 2-Ohm resistor:
- From part (a):
\[
I_2 = 6A
\]

Summary:
- Equivalent resistance: \( 0.545\Omega \)
- Total current: \( 22.02A \)
- Current through the 2-Ohm resistor: \( 6A \)

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#### Problem 4:
Consider the circuit shown at right.

(a) If an ammeter were placed in series beside the 6V battery, what current (I) would it measure?

Circuit Diagram:
- Resistors: \( 3\Omega \), \( 1\Omega \), \( 2\Omega \)
- Battery: \( 6V \)

Solution:
1. Identify the structure:
- All resistors are in series.

2. Total resistance:
\[
R_{\text{total}} = R_1 + R_2 + R_3 = 3\Omega + 1\Omega + 2\Omega = 6\Omega
\]

3. Total current:
\[
I = \frac{V}{R_{\text{total}}} = \frac{6V}{6\Omega} = 1A
\]

Answer:
\[
I = 1A
\]

(b) If a voltmeter measured the potential drop (\( \Delta V \)) across the 1-Ohm resistor, what would it read?

Solution:
1. Potential drop across a resistor:
- Using Ohm's Law:
\[
\Delta V = I \cdot R
\]
- Current \( I \) through the circuit is \( 1A \) (from part (a)).
- Resistance of the 1-Ohm resistor:
\[
R = 1\Omega
\]

2. Calculate the potential drop:
\[
\Delta V = I \cdot R = 1A \cdot 1\Omega = 1V
\]

Answer:
\[
\Delta V = 1V
\]

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#### Problem 5:
The electrical device (D) whose symbol is shown at right requires a terminal voltage of 12V and a current of 2.0A for proper operation. Using only this device and one or more 3Ω resistors, design a circuit so that the device will operate properly when the circuit is connected across a 24V battery.

Solution:
1. Requirements:
- Device (D) needs \( V_D = 12V \) and \( I_D = 2.0A \).
- Battery voltage: \( V_{\text{battery}} = 24V \).

2. Design the circuit:
- To reduce the battery voltage to \( 12V \) for the device, we need a voltage divider.
- Use a single resistor \( R \) in series with the device (D).

3. Calculate the required resistor value:
- Voltage drop across the resistor \( R \):
\[
V_R = V_{\text{battery}} - V_D = 24V - 12V = 12V
\]
- Current through the circuit:
\[
I = I_D = 2.0A
\]
- Using Ohm's Law for the resistor:
\[
R = \frac{V_R}{I} = \frac{12V}{2.0A} = 6\Omega
\]

4. Use available resistors:
- We have 3Ω resistors. To achieve \( 6\Omega \), connect two 3Ω resistors in series:
\[
R_{\text{total}} = 3\Omega + 3\Omega = 6\Omega
\]

5. Final circuit:
- Connect the device (D) in series with two 3Ω resistors in series.
- The total voltage drop across the resistors will be \( 12V \), leaving \( 12V \) for the device.

Circuit Design:
- Device (D) in series with two 3Ω resistors.

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Final Answers:

1. \( I_3 > I_1 = I_2 \)
2. (a) \( A > B = C \)
(b) Bulb A stays the same, bulbs B and C go out.
3. (a) \( I_1 > I_2 > I_3 \)
(b) Equivalent resistance: \( 0.545\Omega \), Total current: \( 22.02A \), Current through 2-Ohm resistor: \( 6A \)
4. (a) \( I = 1A \)
(b) \( \Delta V = 1V \)
5. Use two 3Ω resistors in series with the device.

\boxed{I_3 > I_1 = I_2, A > B = C, I_1 > I_2 > I_3, I = 1A, \Delta V = 1V, \text{Two 3Ω resistors in series}}