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The resistances in the two arms of the meter bridge are 5 $\Omega$ and R $\Omega$, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at $1.6l_1$. The resistance 'R' is:
A
10 $\Omega$
B
15 $\Omega$
C
20 $\Omega$
D
25 $\Omega$
Detailed Solution
$\frac{5}{R} = \frac{l_1}{100 - l_1}$ and $\frac{5}{R/2} = \frac{1.6l_1}{100 - 1.6l_1}$
Solving, $l_1 = 25$ cm and $R = 15\ \Omega$
Solving, $l_1 = 25$ cm and $R = 15\ \Omega$