Current Electricity

JEE Main level, single correct, medium

279 of 380

Question 279 of 380

A meter bridge with two resistances R1R_{1} and R2R_{2} as shown in figure was balanced (null point) at 40 cm from the point PP. The null point changed to 50 cm from the point PP, when 16Ω16 \Omega resistance is connected in parallel to R2R_{2}. The values of resistances R1R_{1} and R2R_{2} are ____\_\_\_\_.

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Answer: (D) R2=8Ω,R1=163ΩR_{2}=8 \Omega, R_{1}=\frac{16}{3} \Omega

Initial balance at 40 cm: R1R2=4060=23\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} ...(i)
With 16 Ω16\,\Omega parallel to R2R_2, null point at 50 cm:
R116R216+R2=5050=1⇒R1=16R216+R2\frac{R_1}{\frac{16R_2}{16+R_2}} = \frac{50}{50} = 1 \Rightarrow R_1 = \frac{16R_2}{16+R_2} ...(ii)
From (i): R1=23R2R_1 = \frac{2}{3}R_2.
Substituting in (ii): 23R2=16R216+R2\frac{2}{3}R_2 = \frac{16R_2}{16+R_2}.
23=1616+R2⇒2(16+R2)=48⇒R2=8 Ω\frac{2}{3} = \frac{16}{16+R_2} \Rightarrow 2(16+R_2) = 48 \Rightarrow R_2 = 8\,\Omega.
R1=163 ΩR_1 = \frac{16}{3}\,\Omega.

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