The current sensitivity of a galvanometer increases by 20%. If its resistance also increases by 25%, the voltage sensitivity will
(i ) decrease by 1%
(ii) increased by 5%
(iii) increased by 10%
(iv ) decrease by 4%
(ii) increased by 5%
(iii) increased by 10%
(iv ) decrease by 4%
(iv ) decrease by 4%
Explanation:
=`\frac{20}{100}"Ig"`
=1.2 `"I'_{"g"}`
R'=R + `"25"/"100"`R
=`"125"/"100"`R
=1.25 R
`"V"'_{"g"}`=?
`"V'"_{"g"}`=`\frac{"I"'_{"g"}}{"R"'}`
=`\frac{"1.2 I"'_{"g"}}{"1.25 R"'}`
=`\frac{120}{125}`Vg=`\frac{25}{25}`Vg
Calculation for % changes
=`\frac{"V"'_{"g"}-"V"_{"g"}}{"V"_{"g"}}\times100`
=`\frac{\frac{25}{24}"V"'_{"g"}"-V"_{"g"}}{"V"_{"g"}}\times100`
=`\frac{24-25}{25}\times100`
=`\frac{-1}{25}\times100`=4%
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