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CGP EDU Academic Team
Published on: September 12, 2026
Two resistances are joined in parallel whose resultant is $\frac{6}{8} \text{ ohm.}$ One of the resistance wire is broken and the effective resistance becomes $2\Omega$ . Then the resistance in ohm of the wire that got broken was
Text Solution
Verified by ExpertsThe correct answer is:
C
If resistances are $R_1$ and $R_2$ then $\frac{R_1 R_2}{R_1 + R_2} = \frac{6}{8}$ ….(i)
Suppose $R_2$ is broken then $R_1 = 2 \Omega$ …(ii)
On solving equations (i) and (ii) we get $R_{2} = 6/5 \Omega$
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