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 ___________.

Solution

Physics sir · 3 steps

Open: two 3R branches in parallel. Closed: the middle nodes merge, so two parallel pairs sit in series.

1. Switch open. The top branch is R + 2R = 3R and the bottom branch is 2R + R = 3R. These two equal branches are in parallel, so Rₒₚₑₙ = 3R/2.
2. Switch closed. The switch shorts the two middle nodes into one point. On the left, R and 2R are in parallel: (R × 2R)/3R = 2R/3. On the right, 2R and R also give 2R/3. These two halves are in series, so R_(closed) = 4R/3.

   Rₒₚₑₙ/R_(closed) = (3R/2)/(4R/3) = 9/8

3. So. The ratio is 9 : 8, so x = 9.

The common slip: 8. Treating the network as a balanced bridge, so that the switch changes nothing, gives a ratio of 8 : 8 and x = 8. Here R/2R ≠ 2R/R, so the bridge is not balanced.

The idea to keep. Closing a switch between two nodes makes them one node. Redraw with that single node, then group resistors again. A bridge is balanced only if the ratios on both sides match.

Answer from the NTA final answer key. Solution by PrepWiser.