Question Number 143758 by Eric002 last updated on 17/Jun/21 Commented by Eric002 last updated on 17/Jun/21 $${use}\:{superposition}\:{to}\:{find}\:{io} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 77765 by mathocean1 last updated on 09/Jan/20 $$\mathrm{Hello}\: \\ $$$$\mathrm{please}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{yield} \\ $$$$\mathrm{in}\:\%\:\mathrm{of}\:\mathrm{ohmic}\:\mathrm{conductor}\:\mathrm{in}\:\mathrm{actif}\:\mathrm{circuit}\: \\ $$$$\mathrm{with}\:\mathrm{Battery}\left(\:\mathrm{with}\:\mathrm{his}\:\mathrm{resistance}\right) \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{motor}\left(\:\mathrm{with}\:\mathrm{his}\:\mathrm{resistance}\right) \\ $$$$\mathrm{knowing}\:\:\mathrm{theirs}\:\mathrm{caracteristics}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{intensity}? \\ $$$$\mathrm{I}\:\mathrm{need}\:\mathrm{a}\:\mathrm{formula}. \\…
Question Number 76038 by Crabby89p13 last updated on 22/Dec/19 Answered by JDamian last updated on 22/Dec/19 $$−\frac{\mathrm{3}}{\mathrm{13}}\:\left({A}\right) \\ $$ Commented by Crabby89p13 last updated on…
Question Number 138831 by mey3nipaba last updated on 18/Apr/21 $${Why}\:{is}\:{the}\:{ammeter}\:{always}\:{connected}\:{in}\:{series} \\ $$$${and}\:{the}\:{voltmeter}\:{in}\:{parallel}? \\ $$ Answered by TheSupreme last updated on 19/Apr/21 $${series}\:{connection}\:{have}\:{same}\:{current}\:\left({ammeter}\:{messure}\:{current}\right) \\ $$$${parallel}\:{connection}\:{saves}\:{ddp}\:\left({voltmeter}\:{measure}\:{ddp}\right) \\…
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Question Number 70746 by Rio Michael last updated on 07/Oct/19 Commented by Rio Michael last updated on 07/Oct/19 $${from}\:{the}\:{above}\:{circuit} \\ $$$$\left.{a}\right){calculate}\:{the}\:{current}\:{flowing}\:{through}\:{the}\:{resistor}\:{R} \\ $$$$\left.{b}\right)\:{find}\:{the}\:{value}\:{of}\:{R} \\ $$$$\left.{c}\right)\:{find}\:{the}\:{emf}\:{E}\:…
Question Number 70568 by Rio Michael last updated on 05/Oct/19 Commented by Rio Michael last updated on 05/Oct/19 $${please}\:{help}. \\ $$$$ \\ $$$${find}\:{from}\:{the}\:{above}\:{circuit},\: \\ $$$$…
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Question Number 69901 by Rio Michael last updated on 28/Sep/19 Commented by Tony Lin last updated on 30/Sep/19 $$\left(\mathrm{1}\right){V}={I}\left({R}_{{A}} +{R}\right) \\ $$$$\because{so}\:{we}\:{may}\:{use}\:{the}\:{first}\:{circuit}\:{to} \\ $$$${measure}\:{something}\:{with}\:{high}\:{resistence} \\…
Question Number 69866 by Askash last updated on 28/Sep/19 Answered by mr W last updated on 28/Sep/19 Commented by mr W last updated on 28/Sep/19…