Question Number 69836 by Askash last updated on 28/Sep/19 Commented by JDamian last updated on 28/Sep/19 $$\mathrm{6}\:\Omega \\ $$ Commented by Askash last updated on…
Question Number 135209 by Eric002 last updated on 11/Mar/21 Commented by Eric002 last updated on 11/Mar/21 $${find}\:{v}\left({ab}\right)\:{by}\:{using}\:{nodal}\:{analysis} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 134520 by Eric002 last updated on 04/Mar/21 Commented by Eric002 last updated on 04/Mar/21 $${find}\:{the}\:{voltage}\:{at}\:{the}\:{nodes}\:{by}\:{using} \\ $$$${nodal}\:{analysis} \\ $$ Terms of Service Privacy…
Question Number 134370 by Eric002 last updated on 03/Mar/21 Commented by Eric002 last updated on 02/Mar/21 $${find}\:{io}\:{and}\:{vo} \\ $$ Answered by Eric002 last updated on…
Question Number 68808 by aseer imad last updated on 15/Sep/19 Answered by JDamian last updated on 15/Sep/19 $${I}=\frac{{V}}{{R}}=\frac{\mathrm{10}}{\mathrm{1}+\mathrm{3}+\mathrm{6}}=\mathrm{1}\:{A} \\ $$$$ \\ $$$${E}_{{c}} =\frac{\mathrm{1}}{\mathrm{2}}{CV}^{\mathrm{2}} \\ $$$$…
Question Number 133765 by Salman_Abir last updated on 24/Feb/21 Answered by JDamian last updated on 24/Feb/21 $${R}_{{e}} =\infty \\ $$ Terms of Service Privacy Policy…
Question Number 68062 by aseer imad last updated on 04/Sep/19 Commented by mind is power last updated on 04/Sep/19 $$\mathrm{2}=\frac{\mathrm{5}}{\mathrm{3}}+\frac{\mathrm{5}}{{R}}\Rightarrow{R}=\mathrm{15}\Omega \\ $$$${E}=\mathrm{10}.\mathrm{2}+\mathrm{2}.\mathrm{2}+\left(\frac{\mathrm{3}.\mathrm{15}}{\mathrm{3}+\mathrm{15}}\right).\mathrm{2}=\mathrm{24}+\mathrm{5}=\mathrm{29}{v} \\ $$ Terms…
Question Number 133104 by Eric002 last updated on 18/Feb/21 Commented by Eric002 last updated on 18/Feb/21 $${find}\:{Vx}\:{by}\:{mesh}\:{analysis} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 132951 by Eric002 last updated on 17/Feb/21 Commented by Eric002 last updated on 17/Feb/21 $${find}\:{Vab} \\ $$ Answered by JDamian last updated on…
Question Number 66249 by Umar last updated on 11/Aug/19 $$\mathrm{The}\:\mathrm{npn}\:\mathrm{transistor}\:\mathrm{in}\:\mathrm{the}\:\mathrm{voltage}\: \\ $$$$\mathrm{amplifier}\:\mathrm{circuit}\:\mathrm{operates}\:\mathrm{satisfctorily} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{quiescent}\:\mathrm{collector}\:\mathrm{current}\:\mathrm{of}\:\mathrm{3mA}. \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{battery}\:\mathrm{supply}\:\left(\mathrm{V}_{\mathrm{cc}} \right)\:\mathrm{is}\:\mathrm{6v},\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:; \\ $$$$\left(\mathrm{a}\right)\:\mathrm{the}\:\mathrm{load}\:\mathrm{resistor}\:\mathrm{R}_{\mathrm{L}} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{the}\:\mathrm{base}\:\mathrm{current}\:\mathrm{for}\:\mathrm{the}\:\mathrm{quiescent} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{collector}−\mathrm{emitter}\:\mathrm{voltage}\:\mathrm{V}_{\mathrm{ce}}…