Solved: 12V Series-Parallel Resistor Circuit

Answer

The equivalent resistance of the circuit is $9.22\,\Omega$, and the total current is $1.30\,\text{A}$. The potential difference across the series $5.0\,\Omega$ resistor is $6.51\,\text{V}$, and the potential difference across the $4.0\,\Omega$ resistor (part of the parallel network) is $2.89\,\text{V}$.

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Interpretation of Image

The provided image shows a combination circuit powered by a $12\,\text{V}$ DC source.

• $R_1 = 2.0\,\Omega$ and $R_2 = 5.0\,\Omega$ are in series with each other and the rest of the circuit.
• $R_3 = 4.0\,\Omega$ and $R_4 = 5.0\,\Omega$ are connected in parallel with each other.
• The total equivalent resistance is the sum of the series resistors and the equivalent resistance of the parallel block.

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Explanation

1. Calculate the parallel equivalent resistance ($R_p$) To find the total resistance, we must first simplify the parallel branch containing $R_3$ and $R_4$. $\frac{1}{R_p} = \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{4.0} + \frac{1}{5.0}$ $\frac{1}{R_p} = 0.25 + 0.20 = 0.45\,\text{S}$ $R_p = \frac{1}{0.45} \approx 2.22\,\Omega$ This is the effective resistance of the parallel branch.

2. Calculate the total equivalent resistance ($R_t$) Now, add the series resistors ($R_1$ and $R_2$) to the parallel result ($R_p$). $R_t = R_1 + R_2 + R_p = 2.0 + 5.0 + 2.22 = 9.22\,\Omega$ The total resistance is the sum of all components in the main loop path. ⚠️ This step is required on exams: always show the intermediate parallel calculation ($R_p$).

3. Calculate the total current ($I_t$) Using Ohm's Law for the whole circuit: $I_t = \frac{V_{total}}{R_t} = \frac{12}{9.22} \approx 1.3015\,\text{A}$ Current is found by dividing the source voltage by the total resistance.

4. Calculate the potential difference across the $5.0\,\Omega$ series resistor ($V_2$) The total current flows through $R_2$ because it is in series with the battery. $V_2 = I_t \times R_2 = 1.3015 \times 5.0 = 6.5075\,\text{V} \approx 6.51\,\text{V}$ Potential difference is the product of current and resistance.

5. Calculate the potential difference across the $4.0\,\Omega$ resistor ($V_3$) The potential difference across the parallel branch ($V_p$) is the same for both $R_3$ and $R_4$. $V_3 = V_p = I_t \times R_p = 1.3015 \times 2.222 \approx 2.892\,\text{V}$ The voltage across a parallel branch is the total current entering the branch times its equivalent resistance.

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Final Answer

• a) Equivalent Resistance: $\boxed{9.22\,\Omega}$
• b) Total Current: $\boxed{1.30\,\text{A}}$
• c) Potential difference across series $5.0\,\Omega$: $\boxed{6.51\,\text{V}}$
• d) Potential difference across $4.0\,\Omega$: $\boxed{2.89\,\text{V}}$

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Common Mistakes

• Inverting the Parallel Formula: Students often calculate $\frac{1}{R_p}$ (e.g., $0.45$) and forget to take the reciprocal to find the actual resistance in Ohms.
• Voltage Misconception: Assuming the source voltage ($12\,\text{V}$) is applied to every resistor. In a series-parallel circuit, the voltage is "shared" across segments; you must calculate the specific drop for each section.
• Premature Rounding: Rounding $R_t$ to $9.2$ too early can lead to slight inaccuracies in the final voltage steps. Keep at least 3 decimal places during intermediate steps.