CBSE Class 12-science Answered
Question no 1
Asked by bhagatshailesh854 | 27 Jun, 2019, 08:13: PM
Expert Answer
A resistance R Ω should be connected in series with the 100 Ω of galvanometer so that
when 5V potential difference is measured, current flowing through galvanometer is limited to 5×10-4 A .
Hence 5 / (R + 100) = 5×10-4 ...................(1)
By solving eqn.(1) for R, we get R = 9.9 kΩ
Answered by Thiyagarajan K | 27 Jun, 2019, 09:26: PM
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