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Two resistances are connected in two gaps of Meter Bridge. The balance is 20cm from the zero end. A resistance of 15 ohms is connected in series with the smaller of the two. The null point shifts to 40cm. What is the value of the bigger resistance?(a) 9(b) 18(c) 27(d) 36The question was asked in an interview for internship.My question is from Current Electricity topic in portion Current Electricity of Physics – Class 12

Answer»

Right option is (d) 36

To EXPLAIN: Let P be the SMALLER resistance and Q be the bigger resistance.

First case→ \(\frac {P}{Q} = \frac {20}{80} = \frac {1}{4}\)

Second case→\(\frac {(P+15)}{Q} = \frac {40}{60} = \frac {2}{3}\)

Comparing both→\(\frac {P}{(P + 15)} = \frac {1}{4} \, \TIMES \, \frac {3}{2} = \frac {3}{8}\)

8P = 3P + 45→5P = 45→P = 9 ohms

Therefore, substituting in \(\frac {P}{Q} = \frac {1}{4}\) → \(\frac {9}{Q} = \frac {1}{4}\) →Q = 36 ohms.


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