Search any question & find its solution
Question:
Answered & Verified by Expert
In a meter bridge a $30 \Omega$ resistance is connected in the left gap and a pair of resistances $P$ and $Q$ in the right gap. Measured from the left, the balance point is $37.5 \mathrm{~cm}$, when $P$ and $Q$ are in series and $71.4 \mathrm{~cm}$ they are in parallel. The values of $P$ and $Q$ (in $\Omega$ ) are
Options:
Solution:
1135 Upvotes
Verified Answer
The correct answer is:
30 20
Ist case
$\begin{aligned}
& \frac{30}{P+Q}=\frac{l}{(100-l)} \\
& \frac{30}{P+Q}=\frac{37.5}{(100-37.5)} \\
& \frac{30}{P+Q}=\frac{37.5}{62.5} \\
& P+Q=\frac{30 \times 62.5}{37.5}
\end{aligned}$
IInd case
$\begin{aligned}
\frac{30}{\frac{P Q}{P+Q}} & =\frac{l}{(100-l)} \\
\frac{30(P+Q)}{P Q} & =\frac{71.4}{(100-71.4)} \\
\frac{30 \times 50}{P Q} & =\frac{71.4}{28.6} \\
P Q & =\frac{30 \times 50 \times 28.6}{71.4}
\end{aligned}$
So, from Eqs. (i) and (ii), we get
$P=30 \Omega, Q=20 \Omega$
$\begin{aligned}
& \frac{30}{P+Q}=\frac{l}{(100-l)} \\
& \frac{30}{P+Q}=\frac{37.5}{(100-37.5)} \\
& \frac{30}{P+Q}=\frac{37.5}{62.5} \\
& P+Q=\frac{30 \times 62.5}{37.5}
\end{aligned}$
IInd case
$\begin{aligned}
\frac{30}{\frac{P Q}{P+Q}} & =\frac{l}{(100-l)} \\
\frac{30(P+Q)}{P Q} & =\frac{71.4}{(100-71.4)} \\
\frac{30 \times 50}{P Q} & =\frac{71.4}{28.6} \\
P Q & =\frac{30 \times 50 \times 28.6}{71.4}
\end{aligned}$
So, from Eqs. (i) and (ii), we get
$P=30 \Omega, Q=20 \Omega$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.