Question

The equivalent resistance between \( A \) and \( B \) as shown in the figure is:

• A. 5 Ω
• B. 10 Ω
• C. 20 Ω
• D. 40 Ω

Answer 1

The Correct Option is A

To find the equivalent resistance between points \( A \) and \( B \), we need to first analyze the circuit configuration presented in the diagram. The given circuit involves resistors in both series and parallel arrangements.

1. Recognize that the resistors between \( A \) and \( C \) as well as \( D \) and \( B \) are each 20 kΩ. The resistor between \( C \) and \( B \) is 10 kΩ.
2. Note that the resistor between \( C \) and \( D \) is also 20 kΩ.
3. The resistors between points \( A \) and \( C \) (20 kΩ), \( C \) and \( D \) (20 kΩ), and \( D \) and \( B \) (20 kΩ) form a parallel network with respect to \( C \) and \( B \).

The circuit consists of three resistors: \( 20 \, \text{k}\Omega \), \( 20 \, \text{k}\Omega \), and \( 10 \, \text{k}\Omega \), connected in parallel. The equivalent resistance of resistors in parallel is given by: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}. \] Substitute \( R_1 = 20 \, \text{k}\Omega \), \( R_2 = 20 \, \text{k}\Omega \), and \( R_3 = 10 \, \text{k}\Omega \): \[ \frac{1}{R_{\text{eq}}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{10}. \] Simplify: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{20} + \frac{1}{20} + \frac{2}{20} = \frac{4}{20} = \frac{1}{5}. \] Solve for \( R_{\text{eq}} \): \[ R_{\text{eq}} = 5 \, \text{k}\Omega. \] Final Answer: The equivalent resistance between \( A \) and \( B \) is: \[ \boxed{5 \, \text{k}\Omega}. \]