The ratio of potential difference across AB in the circuit shown for the case (i) when switch S is closed and (ii) when S is open is:
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Always simplify parallel and series resistor combinations before applying Ohm's law. To find potential differences, first determine total current and then multiply with the resistance across the section of interest.
Let's analyze both scenarios:
Scenario (i): Switch S Closed
When switch S is closed, two parallel paths exist between points B and C:
- Upper path: a resistor of \( 3R \)
- Lower path: a series combination of resistors \( 4R \) and \( 2R \), totaling \( 6R \)
The equivalent resistance between B and C is calculated as:
\[
\frac{1}{R_{BC}} = \frac{1}{3R} + \frac{1}{6R} = \frac{2 + 1}{6R} = \frac{3}{6R} = \frac{1}{2R} \Rightarrow R_{BC} = 2R
\]
The total resistance of the circuit is then:
\[
R_{\text{total}} = 2R \ (\text{resistance of AB}) + 2R \ (\text{equivalent resistance of BC}) = 4R
\]
The current flowing through the circuit is:
\[
I = \frac{V}{4R}
\]
The voltage across the AB segment is:
\[
V_{AB}^{(closed)} = I \cdot 2R = \frac{V}{4R} \cdot 2R = \frac{V}{2}
\]
Scenario (ii): Switch S Open
When switch S is open, only the lower path, which is a series combination of \( 4R \) and \( 2R \) (totaling \( 6R \)), is active between points B and C.
The total resistance of the circuit is:
\[
R_{\text{total}} = 2R \ (\text{resistance of AB}) + 6R \ (\text{resistance of BC}) = 8R
\]
The current flowing through the circuit is:
\[
I = \frac{V}{8R}
\]
The voltage across the AB segment is:
\[
V_{AB}^{(open)} = I \cdot 2R = \frac{V}{8R} \cdot 2R = \frac{V}{4}
\]
Required Ratio:
The ratio of the voltage across AB when the switch is closed to when it is open is:
\[
\frac{V_{AB}^{(closed)}}{V_{AB}^{(open)}} = \frac{\frac{V}{2}}{\frac{V}{4}} = 2
\]
