Question

In the given circuit, the equivalent resistance between points A and D is: 

• A. \( 20 \, \Omega \)
• B. \( 25 \, \Omega \)
• C. \( 35 \, \Omega \)
• D. \( 40 \, \Omega \)

Hint:

When solving circuit problems:
- Identify series and parallel combinations clearly.
- Simplify the circuit step-by-step, reducing it to a single equivalent resistance.
- For parallel resistors, the equivalent resistance is always less than the smallest individual resistance.
- Verify your answer by checking the circuit diagram.

Answer 1

The Correct Option is C

To find the equivalent resistance between points A and D in the given circuit, we analyze the network of resistors step-by-step using series and parallel resistor combinations. This approach simplifies the circuit into a single equivalent resistance.
Step 1: Identify the structure of the circuit
The circuit has two main paths:
- Path A to C: 5 $\Omega$ in series with 10 $\Omega$, followed by 5 $\Omega$ to point C.
- Path B to D: 5 $\Omega$ in series with 10 $\Omega$, followed by 5 $\Omega$ to point D.
- Between C and D: Two 10 $\Omega$ resistors connect C and D in parallel.
Our goal is to simplify this network to find the total resistance between A and D.
Step 2: Simplify the path from A to the first junction
From A to the junction (call it X) where the first 5 $\Omega$ and 10 $\Omega$ meet:
- 5 $\Omega$ and 10 $\Omega$ are in series. \[ R_{\text{AX}} = 5 + 10 = 15 \, \Omega \]
Step 3: Simplify the path from B to the second junction From B to the junction (call it Y) where the first 5 $\Omega$ and 10 $\Omega$ meet: - 5 $\Omega$ and 10 $\Omega$ are in series. \[ R_{\text{BY}} = 5 + 10 = 15 \, \Omega \]
Step 4: Simplify the resistors between X and Y Between points X and Y, there are two 10 $\Omega$ resistors in parallel: \[ \frac{1}{R_{\text{XY}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \] \[ R_{\text{XY}} = 5 \, \Omega \]
Step 5: Include the resistors from X to C and Y to D
- From X to C: 5 $\Omega$ in series.
- From Y to D: 5 $\Omega$ in series.
Now, the path from X to Y includes the parallel combination (5 $\Omega$), and then:
- X to C: 5 $\Omega$.
- Y to D: 5 $\Omega$.
The equivalent resistance from X to Y, including the paths to C and D, needs to be considered carefully. Let’s simplify:
- The 5 $\Omega$ resistors from X to C and Y to D are in series with the parallel combination.
Total resistance from A to D: \[ R_{\text{AX}} + R_{\text{XY}} + R_{\text{YD}} = 15 + 5 + 15 = 35 \, \Omega \]
Step 6: Verify the calculation The simplified circuit is: - 15 $\Omega$ (A to X) in series with 5 $\Omega$ (X to Y, parallel equivalent) in series with 15 $\Omega$ (Y to D). \[ R_{\text{AD}} = 15 + 5 + 15 = 35 \, \Omega \] This confirms our computation.
Step 7: Match with options The options are 20 $\Omega$, 25 $\Omega$, 35 $\Omega$, and 40 $\Omega$. Our calculated value of 35 $\Omega$ matches option (C).
Step 8: Key concepts for beginners
- Series Resistors: Add their resistances directly (\( R_{\text{total}} = R_1 + R_2 \)).
- Parallel Resistors: Use the formula \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \). For two resistors, this simplifies to \( R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \).
- Circuit Simplification: Break the circuit into smaller parts, simplifying series and parallel combinations step-by-step.
- Verification: Always double-check by re-evaluating the circuit structure.