Math Problem Statement

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

Solution

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

Circuit (a):

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

  1. Resistors R1=47ΩR_1 = 47 \, \Omega and R2=82ΩR_2 = 82 \, \Omega are in series: R12=R1+R2=47+82=129ΩR_{12} = R_1 + R_2 = 47 + 82 = 129 \, \Omega

  2. Resistor R3=820ΩR_3 = 820 \, \Omega is in parallel with R12R_{12}: The equivalent resistance for resistors in parallel is given by: 1R123=1R3+1R12=1820+1129\frac{1}{R_{123}} = \frac{1}{R_3} + \frac{1}{R_{12}} = \frac{1}{820} + \frac{1}{129} Solving this gives: R123112.1ΩR_{123} \approx 112.1 \, \Omega

  3. Resistors R4=1.2kΩR_4 = 1.2 \, \text{k}\Omega, R5=91ΩR_5 = 91 \, \Omega, and R6=51ΩR_6 = 51 \, \Omega are in series: R456=R4+R5+R6=1200+91+51=1342ΩR_{456} = R_4 + R_5 + R_6 = 1200 + 91 + 51 = 1342 \, \Omega

  4. Now, R123R_{123} is in series with R456R_{456}: RT=R123+R456=112.1+13421454.1ΩR_T = R_{123} + R_{456} = 112.1 + 1342 \approx 1454.1 \, \Omega

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


Circuit (b):

This is a combination of parallel and series resistors.

  1. Resistors R1=1.2ΩR_1 = 1.2 \, \Omega and R2=2.7ΩR_2 = 2.7 \, \Omega are in series: R12=R1+R2=1.2+2.7=3.9ΩR_{12} = R_1 + R_2 = 1.2 + 2.7 = 3.9 \, \Omega

  2. Resistor R3=3.3ΩR_3 = 3.3 \, \Omega is in parallel with R12R_{12}: The equivalent resistance is: 1R123=1R3+1R12=13.3+13.9\frac{1}{R_{123}} = \frac{1}{R_3} + \frac{1}{R_{12}} = \frac{1}{3.3} + \frac{1}{3.9} Solving this gives: R1231.77ΩR_{123} \approx 1.77 \, \Omega

  3. Finally, R123R_{123} is in series with R4=8.2ΩR_4 = 8.2 \, \Omega: RT=R123+R4=1.77+8.2=9.97ΩR_T = R_{123} + R_4 = 1.77 + 8.2 = 9.97 \, \Omega

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


Summary:

  • RTR_T for circuit (a): 1454.1 Ω\Omega
  • RTR_T for circuit (b): 9.97 Ω\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.

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Math Problem Analysis

Mathematical Concepts

Electricity
Circuits
Resistors in Series and Parallel

Formulas

Resistors in series: R_total = R1 + R2 + ...
Resistors in parallel: 1/R_total = 1/R1 + 1/R2 + ...

Theorems

Ohm's Law
Resistor Combination Rules

Suitable Grade Level

Grades 10-12