Question

In the combinations of resistors shown below, calculate:
(a) the resistance across AB when the switch S is open.
(b) the resistance across AB when the switch S is closed.

Hint:

When analyzing complex resistor networks, first simplify series and parallel combinations. Closing a switch can completely change the circuit's topology, often turning series paths into parallel ones, as seen in this Wheatstone bridge example.

Answer 1

Step 1: Understanding the Concept:
When a switch is open, the branch containing it does not conduct current.
Step 2: Detailed Explanation:
If switch S is open, the circuit consists of two parallel branches:
Branch 1 (Top): \(12 \text{ \(\Omega\)}\) and \(6 \text{ \(\Omega\)}\) in series. \(R_{1} = 12 + 6 = 18 \text{ \(\Omega\)}\).
Branch 2 (Bottom): \(6 \text{ \(\Omega\)}\) and \(12 \text{ \(\Omega\)}\) in series. \(R_{2} = 6 + 12 = 18 \text{ \(\Omega\)}\).
Now, \(R_{1}\) and \(R_{2}\) are in parallel:
\[\frac{1}{R_{AB}} = \frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9}\]
\(R_{AB} = 9 \text{ \(\Omega\)}\).
Step 3: Final Answer:
The resistance across AB when S is open is 9 \(\Omega\).
(b)
Step 1: Understanding the Concept:
When the switch S is closed, the junctions between the resistors are connected. This reconfigures the series/parallel relationships.
Step 2: Detailed Explanation:
With S closed, the circuit becomes:
1. A parallel pair of \(12 \text{ \(\Omega\)}\) and \(6 \text{ \(\Omega\)}\) on the left.
Left equivalent resistance \(R_{L} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4 \text{ \(\Omega\)}\).
2. A parallel pair of \(6 \text{ \(\Omega\)}\) and \(12 \text{ \(\Omega\)}\) on the right.
Right equivalent resistance \(R_{R} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \text{ \(\Omega\)}\).
3. These two equivalent resistors are now in series:
\(R_{AB} = R_{L} + R_{R} = 4 + 4 = 8 \text{ \(\Omega\)}\).
Step 3: Final Answer:
The resistance across AB when S is closed is 8 \(\Omega\).