First slide

Electrical measuring instruments

Question

The resistance in the two arms of the meter bridge are $5 Ω$ and $R Ω$ respectively. When the resistance ‘R’ is shunted with an equal resistance , the new balance point is at $1.6 l_{1}$ . The resistance ‘R’ is

Easy

A

$15 Ω$
B

$20 Ω$
C

$25 Ω$
D

$10 Ω$

Solution

$I n i t i a l c o n d i t i o n \frac{5}{l_{1}} = \frac{R}{100 - l_{1}} - - - - (1) F i n a l c o n d i t i o n \frac{5}{1.6 l_{1}} = \frac{(\frac{R}{2})}{100 - 1.6 l_{1}} \frac{R}{1.6 (100 - l_{1})} = \frac{R}{2 (100 - 1.6 l_{1})} - - - - (2) f r o m e q u a t i o n (1) 1.6 (100 - l_{1}) = 2 (100 - 1.6 l_{1}) \Rightarrow 160 - 1.6 l_{1} = 200 - 3.2 l_{1} \Rightarrow 1.6 l_{1} = 40 \Rightarrow l_{1} = \frac{400}{16} = 25 F r o m e q u a t i o n (1) \frac{5}{25} = \frac{R}{100 - 25} \Rightarrow R = 15 Ω$

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