Question

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) 36

The question was asked in an interview for internship.

My question is from Current Electricity topic in portion Current Electricity of Physics – Class 12

Answer 1

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.