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 cases:
Case (i): Switch S Closed
When switch S is closed, two branches from B to C:
- Upper path: resistor \( 3R \)
- Lower path: series combination \( 4R + 2R = 6R \)
These two are in parallel, so equivalent resistance:
\[
\frac{1}{R_{BC}} = \frac{1}{3R} + \frac{1}{6R} = \frac{2 + 1}{6R} = \frac{3}{6R} = \frac{1}{2R} \Rightarrow R_{BC} = 2R
\]
Now the total resistance in the circuit:
\[
R_{\text{total}} = 2R \ (\text{AB}) + 2R \ (\text{BC}) = 4R
\]
Current:
\[
I = \frac{V}{4R}
\]
Voltage across AB:
\[
V_{AB}^{(closed)} = I \cdot 2R = \frac{V}{4R} \cdot 2R = \frac{V}{2}
\]
Case (ii): Switch S Open
Only the lower path \( 4R + 2R = 6R \) is active:
\[
R_{\text{total}} = 2R + 6R = 8R
\]
Current:
\[
I = \frac{V}{8R}
\]
Voltage across AB:
\[
V_{AB}^{(open)} = I \cdot 2R = \frac{V}{8R} \cdot 2R = \frac{V}{4}
\]
Required Ratio:
\[
\frac{V_{AB}^{(closed)}}{V_{AB}^{(open)}} = \frac{\frac{V}{2}}{\frac{V}{4}} = 2
\]
