https://file.aiquickdraw.com/mathbot/file/1726696266957hzu4po9c.jpg

Let's analyze both circuits shown in the image to calculate the total resistance (\( R_T \)) for each one.

### Circuit (a):

This circuit consists of a combination of resistors in both series and parallel.

1. **Resistors \(R_1 = 47 \, \Omega\) and \(R_2 = 82 \, \Omega\) are in series:**
   \[
   R_{12} = R_1 + R_2 = 47 + 82 = 129 \, \Omega
   \]

2. **Resistor \(R_3 = 820 \, \Omega\) is in parallel with \(R_{12}\):**
   The equivalent resistance for resistors in parallel is given by:
   \[
   \frac{1}{R_{123}} = \frac{1}{R_3} + \frac{1}{R_{12}} = \frac{1}{820} + \frac{1}{129}
   \]
   Solving this gives:
   \[
   R_{123} \approx 112.1 \, \Omega
   \]

3. **Resistors \(R_4 = 1.2 \, \text{k}\Omega\), \(R_5 = 91 \, \Omega\), and \(R_6 = 51 \, \Omega\) are in series:**
   \[
   R_{456} = R_4 + R_5 + R_6 = 1200 + 91 + 51 = 1342 \, \Omega
   \]

4. **Now, \(R_{123}\) is in series with \(R_{456}\):**
   \[
   R_T = R_{123} + R_{456} = 112.1 + 1342 \approx 1454.1 \, \Omega
   \]

So, the total resistance for circuit (a) is approximately **1454.1 \(\Omega\)**.

---

### Circuit (b):

This is a combination of parallel and series resistors.

1. **Resistors \(R_1 = 1.2 \, \Omega\) and \(R_2 = 2.7 \, \Omega\) are in series:**
   \[
   R_{12} = R_1 + R_2 = 1.2 + 2.7 = 3.9 \, \Omega
   \]

2. **Resistor \(R_3 = 3.3 \, \Omega\) is in parallel with \(R_{12}\):**
   The equivalent resistance is:
   \[
   \frac{1}{R_{123}} = \frac{1}{R_3} + \frac{1}{R_{12}} = \frac{1}{3.3} + \frac{1}{3.9}
   \]
   Solving this gives:
   \[
   R_{123} \approx 1.77 \, \Omega
   \]

3. **Finally, \(R_{123}\) is in series with \(R_4 = 8.2 \, \Omega\):**
   \[
   R_T = R_{123} + R_4 = 1.77 + 8.2 = 9.97 \, \Omega
   \]

Thus, the total resistance for circuit (b) is approximately **9.97 \(\Omega\)**.

---

### Summary:
- \(R_T\) for circuit (a): **1454.1 \(\Omega\)**
- \(R_T\) for circuit (b): **9.97 \(\Omega\)**

Would you like more details on any specific part of the calculation?

Here are five related questions to deepen your understanding:
1. How would you calculate the equivalent resistance of more complex parallel networks?
2. What happens to the total resistance when more resistors are added in parallel?
3. Why is the equivalent resistance always less than the smallest resistor in a parallel configuration?
4. How does the configuration of resistors affect power distribution in the circuit?
5. Can you explain the differences between series and parallel combinations in terms of voltage and current?

**Tip:** When resistors are in parallel, the equivalent resistance is always less than the smallest resistor in the group.

A resistência dos circuitos mostrados a seguir valem: (0.5 Pontos)

Learn how to calculate the total resistance of series and parallel resistor circuits, including step-by-step solutions for two sample circuits. Includes formulas for resistors in series and parallel, and covers key concepts for Grades 10-12.

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