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

(a) Resistance when switch S is open:
When the switch S is open, there is no connection between the upper and lower branches at the center. The circuit consists of two parallel branches between points A and B.
- Upper branch: A 12 \(\Omega\) resistor and a 6 \(\Omega\) resistor are in series.
\(R_{upper} = 12 \, \Omega + 6 \, \Omega = 18 \, \Omega\).
- Lower branch: A 6 \(\Omega\) resistor and a 12 \(\Omega\) resistor are in series.
\(R_{lower} = 6 \, \Omega + 12 \, \Omega = 18 \, \Omega\).
The total resistance across AB (\(R_{AB}\)) is the equivalent resistance of these two branches in parallel.
\[ \frac{1}{R_{AB}} = \frac{1}{R_{upper}} + \frac{1}{R_{lower}} = \frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9} \] \[ R_{AB} = 9 \, \Omega \] (b) Resistance when switch S is closed:
When the switch S is closed, it connects the midpoint of the upper branch to the midpoint of the lower branch. This reconfigures the circuit into two parallel combinations connected in series. This is a Wheatstone bridge configuration.
- Left side of S: The 12 \(\Omega\) resistor and the 6 \(\Omega\) resistor are in parallel.
\[ R_{left} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4 \, \Omega \] - Right side of S: The 6 \(\Omega\) resistor and the 12 \(\Omega\) resistor are in parallel.
\[ R_{right} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \, \Omega \] The total resistance across AB is the sum of these two combinations in series. \[ R_{AB} = R_{left} + R_{right} = 4 \, \Omega + 4 \, \Omega = 8 \, \Omega \]