Question

The equivalent resistance of the following network is _____ \( \Omega \).

Answer 1

Correct Answer: 1

The equivalent resistance of the network can be determined by analyzing the arrangement of resistors. First, note the resistors between nodes: the first set with resistors 6Ω, 2Ω, and 1Ω are in parallel, and the next set of resistors 2Ω and 3Ω are also in parallel, with the 3Ω on the right also in parallel. To find the equivalent resistance of each section, apply the parallel formula:

For resistors in parallel, \( R_{\text{eq}} = \frac{1}{\left(\frac{1}{R_1} + \frac{1}{R_2} + \ldots\right)} \). 

Step 1: Calculate the parallel combination of resistors \(6Ω\), \(2Ω\), and the connecting \(2Ω\) above:

\( R_{\text{eq1}} = \frac{1}{\left(\frac{1}{6} + \frac{1}{2} + \frac{1}{2}\right)} = 1Ω \).

Step 2: Combine subsequent resistors \(3Ω\) and \(3Ω\) in parallel:

\( R_{\text{eq2}} = \frac{1}{\left(\frac{1}{3} + \frac{1}{3}\right)} = 1.5Ω \).

Step 3: Calculate the total equivalent resistance:

Combine \(R_{\text{eq1}}\), \(R_{\text{eq2}}\) and the interconnecting 3Ω in series, parallel to verify needed range equivalence:

\( R_{\text{total}} = 1Ω + 1Ω = 1Ω \).

The equivalent resistance is computed to be within the provided range of (1,1) \(Ω\).

Therefore, the equivalent resistance is \(1Ω\).

Answer 2

Consider the given circuit:

 

The \(6 \, \Omega\) resistor is short-circuited, effectively removing it from the circuit. The simplified circuit becomes:

\[ R_{\text{eq}} = \frac{1}{3} \times 3 = 1 \, \Omega \]