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In a practical wheat stone bridge circuit as shown, when one more resistance of 100 \Omega is connected in parallel with unknown resistance ' x ', then ratio l _1 / l_2 become ' 2 '. l _1 is balance length. AB is a uniform wire. Then  value of ' x ' must be:

Option: 1

50 


Option: 2

100 


Option: 3

200 


Option: 4

400 


Answers (1)

best_answer

wheat stone bridge is in balanced condition

            So \frac{100}{l_1} = \frac{\frac{100x}{100+x}}{l_2}

\frac{l_1}{l_2}= 2 \\\\ x = 100 \Omega

Posted by

sudhir.kumar

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