When the right gap of a metre bridge consists of two equal resistors in series, the balancing point is at 50 cm . When one of the resisters in the right gap is removed and is connected in parallel to the resistor in the left gap, the balancing point is at
First setup: Let the resistance in the left gap be R′, and in the right gap there are two resistors with resistance R, connected in series. So, the right gap has a total resistance of R+R=2R. The balancing point is at 50 cm . In a metre bridge, the ratio of resistances equals the ratio of the lengths on both sides: ‌R′∕2R=‌50∕50 So, ‌R′∕2R=1 which means R′=2R Second setup: Now, one resistor from the right gap is taken out and connected in parallel with the resistor in the left gap. Now, the left gap contains R′ and R in parallel. The resistance of two resistors A and B in parallel is: R′′=‌A×B∕A+B The parallel resistance is: R′′=‌R′×R∕R′+R From earlier, we know R′=2R. Substitute this value: R′′=‌2R×R∕2R+R=‌2R2∕3R=‌2R∕3 Now, the right gap just has R. Let the new balancing point be l. The bridge principle gives us: ‌R′′∕R=‌l∕100−l Substitute R′′=‌2R∕3 : ‌‌2R∕3∕R=‌l∕100−l‌2∕3=‌l∕100−l Solve for l:3l=2(100−l)3l=200−2l3l+2l=2005l=200l=40cm