The circuit shown in the image has resistors arranged in a combination of series and parallel.

1. Series combination of resistors: When resistors are connected in series, the total (or equivalent) resistance is the sum of the resistances: $R_{\text{series}} = R_1 + R_2$

2. Parallel combination of resistors: When resistors are connected in parallel, the total (or equivalent) resistance is calculated using the formula: $\frac{1}{R_{\text{parallel}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3}$ Or equivalently: $R_{\text{parallel}} = \frac{R_{\text{series}} \cdot R_3}{R_{\text{series}} + R_3}$

Step-by-Step Solution:

• Given:
  • $R_1 = 43 \, \Omega$
  • $R_2 = 66 \, \Omega$
  • $R_3 = 63 \, \Omega$

1. Combine $R_1$ and $R_2$ in series: $R_{\text{series}} = R_1 + R_2 = 43 \, \Omega + 66 \, \Omega = 109 \, \Omega$

2. Combine the result with $R_3$ in parallel: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3} = \frac{1}{109 \, \Omega} + \frac{1}{63 \, \Omega}$ Simplifying this: $\frac{1}{R_{\text{eq}}} = 0.009174 + 0.015873 = 0.025047$ Thus, the equivalent resistance is: $R_{\text{eq}} = \frac{1}{0.025047} \approx 39.92 \, \Omega$

   Rounding to two significant figures: $R_{\text{eq}} \approx 40 \, \Omega$

Final Answer: $R_{\text{eq}} = 40 \, \Omega$

Would you like further details on this? Here are some related questions:

1. What is the difference between resistors in series and in parallel?
2. How would the equivalent resistance change if the value of $R_3$ were halved?
3. What would be the impact of increasing $R_1$ by 20% on the total resistance?
4. Can you explain the formula for resistors in parallel more intuitively?
5. How is energy dissipated in such a resistor network?

Tip: Always remember to check whether resistors are in series or parallel before calculating the equivalent resistance in a circuit.