Worksheet B: Combination Circuits - Question 1 asks to rank currents I1 through I3 from largest to smallest.

Here are the step-by-step solutions for the problems on your worksheet.

Problem 1: Ranking Currents in a Combination Circuit

Step 1: Analyze the circuit structure.
The current $I_1$ comes out of the battery and goes through resistor $R_1$. After passing through $R_1$, the wire splits into two branches. One branch has resistor $R_2$ (carrying current $I_2$) and the other has resistor $R_3$ (carrying current $I_3$). Finally, these branches recombine to return to the battery.

Step 2: Apply Kirchhoff's Current Law.
The total current entering a junction must equal the total current leaving it. Therefore, the main current $I_1$ splits into $I_2$ and $I_3$.
Equation: $I_1 = I_2 + I_3$.
Because $I_1$ is the sum of the other two, $I_1$ is definitely the largest current.

Step 3: Compare $I_2$ and $I_3$.
Current follows the path of least resistance. In parallel branches with the same voltage across them, the branch with lower resistance gets more current.
* Resistance of branch 2 ($R_2$) = $R$
* Resistance of branch 3 ($R_3$) = $2R$

Since $R_2$ is smaller than $R_3$ (half the resistance), twice as much current will flow through it compared to $R_3$.
Therefore, $I_2 > I_3$.

Conclusion:
Order: $I_1 > I_2 > I_3$

Explanation: $I_1$ is the total current before the split, so it is the largest. Between the parallel branches, current is inversely proportional to resistance. Since $R_2$ ($R$) is less than $R_3$ ($2R$), more current flows through $R_2$.

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Problem 2: Brightness of Bulbs

Part a) Rank brightness

Step 1: Understand Brightness.
Brightness depends on power ($P = I^2R$ or $P = V^2/R$). Since the bulbs are identical, they have the same resistance ($R$). We just need to compare the current flowing through them or the voltage across them.

Step 2: Analyze Current Flow.
Current leaves the battery and goes entirely through Bulb A. So, $I_A$ is the total circuit current.
After Bulb A, the current splits between Bulb B and Bulb C.
Equation: $I_A = I_B + I_C$. This means Bulb A is the brightest because it has the most current.

Step 3: Compare Bulb B and Bulb C.
Look at the diagram for Bulbs B and C. They are connected in parallel to each other.
* The top of Bulb B and the top of Bulb C connect to the same wire.
* The bottom of Bulb B and the bottom of Bulb C connect to the same wire.
Because they are in parallel, they have the exact same voltage across them. Since they are identical bulbs, they will draw the exact same current.
Therefore, Bulb B and Bulb C have equal brightness.

Order: A > B = C

Explanation: All current passes through A, making it the brightest. The current then splits equally between the identical parallel bulbs B and C, making them dimmer than A but equal to each other.

Part b) Connecting a wire between points 1 and 2

Step 1: Identify Points 1 and 2.
Point 1 is on the wire connecting the top of Bulb C to the rest of the circuit. Point 2 is on the wire connecting the bottom of Bulb C to the rest of the circuit. Essentially, points 1 and 2 are the two ends of Bulb C.

Step 2: Effect of the Wire (Short Circuit).
Connecting a plain wire between points 1 and 2 creates a "short circuit" across Bulb C. Electricity prefers to flow through a wire (zero resistance) rather than a light bulb (high resistance). Therefore, almost all current bypasses Bulb C.
* Bulb C: Goes out (no current flows through it).

Step 3: Effect on Total Resistance and Current.
Originally, the resistance of the parallel part (B and C) was $R/2$ (since two identical resistors in parallel halve the resistance).
Now that C is shorted out, the current only flows through Bulb B in that section. The resistance of that section becomes just $R$ (the resistance of Bulb B).
The total resistance of the entire circuit increases (from $R_A + 0.5R$ to $R_A + R$).

Step 4: Effect on Bulbs A and B.
Since the total resistance of the circuit increased, the total current coming from the battery ($I_{total}$) decreases.
* Bulb A: Since $I_{total}$ flows through A, and $I_{total}$ decreased, Bulb A gets dimmer.
* Bulb B: This is tricky. Originally, Bulb B got half of the total current. Now, it gets *all* of the new (smaller) total current.
* Old Current in B: $0.5 \times I_{old\_total}$
* New Current in B: $1.0 \times I_{new\_total}$
* Let's check the math:
* Old Total Resistance = $1.5 R$. Old Total Current = $V / 1.5R = 0.67 V/R$. Current in B = $0.33 V/R$.
* New Total Resistance = $2.0 R$. New Total Current = $V / 2.0R = 0.5 V/R$. Current in B = $0.5 V/R$.
* Since $0.5 > 0.33$, the current in Bulb B actually increases.
* Bulb B gets brighter.

Answer:
* Bulb A: Gets dimmer.
* Bulb B: Gets brighter.
* Bulb C: Goes out.

Explanation: The wire shorts out Bulb C, removing it from the circuit. This increases the total resistance of the circuit, which lowers the total current, making Bulb A dimmer. However, Bulb B now receives the full total current instead of sharing it, and this new share is larger than its previous share, so it gets brighter.

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Problem 3: Currents Through Parallel Circuit

Part a) Rank currents

Step 1: Analyze Voltage.
In a parallel circuit, the voltage across every branch is the same as the battery voltage. So, $V_1 = V_2 = V_3 = 12V$.

Step 2: Use Ohm's Law ($I = V/R$).
Since Voltage ($V$) is constant, Current ($I$) is inversely proportional to Resistance ($R$). The smaller the resistor, the bigger the current.
* Resistor 1: $1 \Omega$ (Smallest R $\rightarrow$ Largest I)
* Resistor 2: $2 \Omega$
* Resistor 3: $3 \Omega$ (Largest R $\rightarrow$ Smallest I)

Order: $I_1 > I_2 > I_3$

Explanation: In parallel circuits, voltage is the same across all branches. According to Ohm's law, current is inversely proportional to resistance. Therefore, the branch with the least resistance ($1\Omega$) carries the most current, and the branch with the most resistance ($3\Omega$) carries the least.

Part b) Calculate Equivalent Resistance and Currents

Step 1: Calculate Equivalent Resistance ($R_{eq}$).
Formula for parallel resistors:
$$ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$
$$ \frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} $$
Find a common denominator (which is 6):
$$ \frac{1}{R_{eq}} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} $$
$$ \frac{1}{R_{eq}} = \frac{11}{6} $$
Flip the fraction to solve for $R_{eq}$:
$$ R_{eq} = \frac{6}{11} \Omega \approx 0.545 \Omega $$

Step 2: Calculate Total Current ($I_{total}$).
$$ I_{total} = \frac{V}{R_{eq}} = \frac{12}{\frac{6}{11}} = 12 \times \frac{11}{6} $$
$$ I_{total} = 2 \times 11 = 22 \text{ Amps} $$

Step 3: Calculate Current through the 2-Ohm Resistor ($I_2$).
We know the voltage across this specific resistor is 12V.
$$ I_2 = \frac{V}{R_2} = \frac{12V}{2\Omega} = 6 \text{ Amps} $$

Final Answer for 3b:
Equivalent Resistance: $6/11 \Omega$ (or approx $0.55 \Omega$)
Total Current: $22 \text{ A}$
Current through 2-Ohm resistor: $6 \text{ A}$

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Problem 4: Ammeter and Voltmeter Readings

Circuit Analysis:
* Battery: 6V
* Resistor 1: $3\Omega$ (in series with the rest)
* Parallel Block: $1\Omega$ resistor || $2\Omega$ resistor

Step 1: Find Total Resistance.
First, find the equivalent resistance of the parallel part ($R_p$):
$$ \frac{1}{R_p} = \frac{1}{1} + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2} $$
$$ R_p = \frac{2}{3} \Omega $$

Now add the series resistor ($3\Omega$):
$$ R_{total} = 3 + \frac{2}{3} = \frac{9}{3} + \frac{2}{3} = \frac{11}{3} \Omega $$

Step 2: Find Total Current ($I_{total}$).
$$ I_{total} = \frac{V}{R_{total}} = \frac{6}{\frac{11}{3}} = 6 \times \frac{3}{11} = \frac{18}{11} \text{ Amps} \approx 1.64 \text{ A} $$

Part a) Ammeter beside the battery
An ammeter measures the current flowing through the wire it is placed in. Beside the battery, it measures the total current leaving the battery.
Reading: $\frac{18}{11} \text{ A}$ (or approx $1.64 \text{ A}$)

Part b) Voltmeter across the $1\Omega$ resistor
The voltmeter measures the voltage drop across the parallel combination.
We can calculate this by subtracting the voltage drop of the first resistor ($3\Omega$) from the total battery voltage, or by calculating $V = I_{total} \times R_{parallel}$.

Method: $V_p = I_{total} \times R_p$
$$ V_p = \left(\frac{18}{11}\right) \times \left(\frac{2}{3}\right) $$
$$ V_p = \frac{36}{33} $$
Simplify by dividing top and bottom by 3:
$$ V_p = \frac{12}{11} \text{ Volts} \approx 1.09 \text{ V} $$

Reading: $\frac{12}{11} \text{ V}$ (or approx $1.09 \text{ V}$)

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Problem 5: Designing a Circuit

Goal:
* Device D needs: 12V and 2.0A.
* Source: 24V Battery.
* Available parts: Device D, one or more $3\Omega$ resistors.

Step 1: Determine the requirement for the Device.
Using Ohm's Law, we can find the internal resistance of Device D, though it's not strictly necessary if we think about voltage drops.
$R_{device} = \frac{V}{I} = \frac{12V}{2A} = 6\Omega$.

Step 2: Analyze Voltage Needs.
The battery provides 24V. The device only wants 12V.
This means we need to "drop" the remaining 12V somewhere else in the circuit so that only 12V is left for the device.
$$ V_{resistor} = V_{battery} - V_{device} = 24V - 12V = 12V $$

Step 3: Analyze Current Needs.
The device draws 2.0A. If we place a resistor in series with the device, the same 2.0A current will flow through that resistor.

Step 4: Calculate Required Resistance.
We need a resistor (or combination of resistors) that will drop 12V when 2.0A flows through it.
$$ R_{needed} = \frac{V_{drop}}{I} = \frac{12V}{2.0A} = 6\Omega $$

Step 5: Construct the Resistor using $3\Omega$ parts.
We need a total resistance of $6\Omega$. We have $3\Omega$ resistors.
If we put two $3\Omega$ resistors in series, their resistances add up:
$$ 3\Omega + 3\Omega = 6\Omega $$

Design Description:
Connect the positive terminal of the 24V battery to one end of a $3\Omega$ resistor. Connect the other end of that resistor to one end of a second $3\Omega$ resistor. Connect the other end of the second resistor to Terminal A of Device D. Connect Terminal B of Device D back to the negative terminal of the battery.

(Essentially: Battery $\rightarrow$ Resistor ($3\Omega$) $\rightarrow$ Resistor ($3\Omega$) $\rightarrow$ Device D $\rightarrow$ Battery)

Final Answer:
Connect two $3\Omega$ resistors in series with Device D.