To solve this problem, let's proceed step by step.

Step 1: Total Resistance of the Circuit

Since the resistors are connected in series, the total resistance $R_{total}$ is the sum of the individual resistances:

$R_{total} = R_1 + R_2 + R_3$ Where:

• $R_1 = 4 \, \Omega$
• $R_2 = 9 \, \Omega$
• $R_3 = 11 \, \Omega$

$R_{total} = 4 \, \Omega + 9 \, \Omega + 11 \, \Omega = 24 \, \Omega$

Step 2: Current Flowing through the Circuit

Using Ohm's Law $V = I \times R$, we can calculate the total current flowing through the circuit. The voltage of the battery $V$ is 12V.

$I = \frac{V}{R_{total}} = \frac{12 \, V}{24 \, \Omega} = 0.5 \, A$

So, the current flowing through the circuit is 0.5 A.

Step 3: Potential Difference across the 9Ω Resistor

The potential difference across a resistor can also be determined using Ohm's Law $V = I \times R$.

For the 9Ω resistor:

$V_{9 \, \Omega} = I \times R_2 = 0.5 \, A \times 9 \, \Omega = 4.5 \, V$

Thus, the potential difference across the 9Ω resistor is 4.5 V.

Step 4: Power Dissipated in the 11Ω Resistor

The power dissipated by a resistor can be calculated using the formula:

$P = I^2 \times R$

For the 11Ω resistor:

$P_{11 \, \Omega} = (0.5 \, A)^2 \times 11 \, \Omega = 0.25 \, A^2 \times 11 \, \Omega = 2.75 \, W$

So, the power dissipated in the 11Ω resistor is 2.75 W.

Summary of Results:

• Current through the circuit: $0.5 \, A$
• Potential difference across the 9Ω resistor: $4.5 \, V$
• Power dissipated in the 11Ω resistor: $2.75 \, W$

Would you like further details or have any questions?

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Here are five related questions to explore further:

1. What happens to the current if an additional resistor is added in series?
2. How would the power dissipated in the 9Ω resistor be calculated?
3. What is the total power dissipated in the entire circuit?
4. How does the current change if the battery voltage increases to 24V?
5. How would you calculate the voltage drop if the resistors were connected in parallel?

Tip: In a series circuit, the current is the same through all components, but the voltage drops vary depending on the resistance.