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Question Number 8571 by tawakalitu last updated on 16/Oct/16

Commented by sandy_suhendra last updated on 16/Oct/16

$$\mathrm{Is}\mathrm{the}\mathrm{potensial}\mathrm{of}\mathrm{supply}1.0\mathrm{volt}?$$

Commented by tawakalitu last updated on 16/Oct/16

$$\mathrm{yes}\mathrm{sir}.$$

Answered by ridwan balatif last updated on 17/Oct/16

Answered by ridwan balatif last updated on 17/Oct/16

Commented by tawakalitu last updated on 17/Oct/16

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by sandy_suhendra last updated on 17/Oct/16

Commented by sandy_suhendra last updated on 17/Oct/16

$$\mathrm{if}\mathrm{the}\mathrm{voltmeter}\mathrm{is}\mathrm{disconnected},\mathrm{the}\mathrm{circuit}\mathrm{looks}\mathrm{like}\mathrm{above}$$$$\mathrm{the}\mathrm{potensial}\mathrm{of}\mathrm{supply}=\mathrm{E}=1\mathrm{volt}$$$$\mathrm{so}\mathrm{\Delta }{\mathrm{V}}_{\mathrm{AB}}=\mathrm{E}=1\mathrm{volt}=\mathrm{\Delta }{\mathrm{V}}_{\mathrm{AC}}+\mathrm{\Delta }{\mathrm{V}}_{\mathrm{CB}}$$$$\mathrm{\Delta }{\mathrm{V}}_{\mathrm{AC}}:\mathrm{\Delta }{\mathrm{V}}_{\mathrm{CB}}={\mathrm{R}}_1:{\mathrm{R}}_2=500:1500=1:3$$$$\mathrm{so}\mathrm{\Delta }{\mathrm{V}}_{\mathrm{CB}}=\frac{3}{4}\times 1=\frac{3}{4}\mathrm{volt}$$$$\mathrm{\Delta }{\mathrm{V}}_{\mathrm{AC}}=\frac{1}{4}\times 1=\frac{1}{4}\mathrm{volt}$$$$(\mathrm{when}\mathrm{the}\mathrm{voltmeter}\mathrm{is}\mathrm{connected},\mathrm{I}\mathrm{think}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{imposible}\mathrm{if}\mathrm{E}=1\mathrm{volt}\mathrm{and}\mathrm{the}\mathrm{p}.\mathrm{d}.\mathrm{reads}\mathrm{at}$$$$\mathrm{the}\mathrm{voltmeter}\mathrm{is}1\mathrm{volt}\mathrm{too})$$

Commented by tawakalitu last updated on 17/Oct/16

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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