Question

The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is x : 8. The value of x is _________. 

 

Hint:

When analyzing complex resistor networks, always look for simplifications. Check for series/parallel combinations first. If a switch is involved, analyze the "open" and "closed" circuits separately. For the closed case here, redrawing the circuit can often clarify the parallel connections.

Answer 1

Correct Answer: 9

Step 1: Understanding the Question:
We need to find the equivalent resistance of the given circuit in two scenarios: first when the switch S is open, and second when it's closed. Then we use the given ratio of these two resistances to find the value of x.
Step 2: Detailed Explanation:
Case 1: Switch S is open (\(R_{\text{open}}\))
When the switch is open, no current flows through it. The circuit consists of two parallel branches.
- Top Branch: Resistor R and 2R are in series. Their combined resistance is \( R_{\text{top}} = R + 2R = 3R \).
- Bottom Branch: Resistor 2R and R are in series. Their combined resistance is \( R_{\text{bottom}} = 2R + R = 3R \).
These two branches are in parallel between points a and b. The equivalent resistance is:
\[ R_{\text{open}} = \frac{R_{\text{top}} \times R_{\text{bottom}}}{R_{\text{top}} + R_{\text{bottom}}} = \frac{(3R) \times (3R)}{3R + 3R} = \frac{9R^2}{6R} = \frac{3}{2}R \] Case 2: Switch S is closed (\(R_{\text{closed}}\))
When the switch is closed, it connects the midpoint of the top and bottom branches. This forms a network that can be seen as two parallel combinations in series.
- Left side: Resistor R is in parallel with resistor 2R. Their equivalent resistance is \( R_{\text{left}} = \frac{R \times 2R}{R + 2R} = \frac{2R^2}{3R} = \frac{2}{3}R \).
- Right side: Resistor 2R is in parallel with resistor R. Their equivalent resistance is \( R_{\text{right}} = \frac{2R \times R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3}R \).
These two combinations are in series between points a and b.
\[ R_{\text{closed}} = R_{\text{left}} + R_{\text{right}} = \frac{2}{3}R + \frac{2}{3}R = \frac{4}{3}R \] Note: This is not a balanced Wheatstone bridge since \(R/(2R) \neq (2R)/R\).
Step 3: Finding the value of x
We are given that the ratio \( \frac{R_{\text{open}}}{R_{\text{closed}}} = \frac{x}{8} \).
Let's compute the ratio using our results:
\[ \frac{R_{\text{open}}}{R_{\text{closed}}} = \frac{\frac{3}{2}R}{\frac{4}{3}R} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} \] Comparing this with the given ratio:
\[ \frac{9}{8} = \frac{x}{8} \] This implies \( x = 9 \).
Step 4: Final Answer:
The value of x is 9.