Question Number 40975 by rahul 19 last updated on 30/Jul/18
Answered by ajfour last updated on 30/Jul/18
$$\left({b}\right)\:\mathrm{10}{A}/{s} \\ $$
Commented by rahul 19 last updated on 31/Jul/18
$$\mathrm{Yes}\:\mathrm{sir}, \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18
![whent=0 L offer infinite resistance so no current through L...hence (5+1=6 ohm),6ohm and (2+4=6ohm in parallel connection (1/R)=(1/6)+(1/6)+(1/6)=(3/6) R=2ohm curednt=(6/2)=3A through AB→BD 1A current thtough A→D 1A current throughA→C→D 1A current...V_A −V_B =1×1 6−V_B =1 so V_B =5V V_A −V_c =2×1=2 V 6−V_c =2 V_c =4V V_B =5V V_c .=4V V_B −V_c =L(di/dt) (di/dt)=((V_B −V_c )/L)=((5−4)/(O.1H))=(/(1/(10)))=10 (di/dt)=((10 A]/second...)/)](Q41004.png)
$${whent}=\mathrm{0}\:{L}\:{offer}\:{infinite}\:{resistance}\:{so}\:{no}\:{current} \\ $$$${through}\:{L}...{hence}\:\left(\mathrm{5}+\mathrm{1}=\mathrm{6}\:{ohm}\right),\mathrm{6}{ohm}\:{and} \\ $$$$\left(\mathrm{2}+\mathrm{4}=\mathrm{6}{ohm}\:{in}\:{parallel}\:{connection}\right. \\ $$$$\frac{\mathrm{1}}{{R}}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{6}}\:\:\:\:{R}=\mathrm{2}{ohm} \\ $$$${curednt}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3}{A} \\ $$$${through}\:{AB}\rightarrow{BD}\:\:\:\mathrm{1}{A}\:{current} \\ $$$${thtough}\:{A}\rightarrow{D}\:\:\:\mathrm{1}{A}\:\:{current} \\ $$$${throughA}\rightarrow{C}\rightarrow{D}\:\:\mathrm{1}{A}\:{current}...{V}_{{A}} −{V}_{{B}} =\mathrm{1}×\mathrm{1} \\ $$$$\mathrm{6}−{V}_{{B}} =\mathrm{1}\:\:\:{so}\:{V}_{{B}} =\mathrm{5}{V} \\ $$$${V}_{{A}} −{V}_{{c}} =\mathrm{2}×\mathrm{1}=\mathrm{2}\:{V} \\ $$$$\mathrm{6}−{V}_{{c}} =\mathrm{2}\:\:\:{V}_{{c}} =\mathrm{4}{V} \\ $$$${V}_{{B}} =\mathrm{5}{V}\:\:\:\:\:{V}_{{c}} .=\mathrm{4}{V} \\ $$$${V}_{{B}} −{V}_{{c}} ={L}\frac{{di}}{{dt}} \\ $$$$\frac{{di}}{{dt}}=\frac{{V}_{{B}} −{V}_{{c}} }{{L}}=\frac{\mathrm{5}−\mathrm{4}}{{O}.\mathrm{1}{H}}=\frac{}{\frac{\mathrm{1}}{\mathrm{10}}}=\mathrm{10} \\ $$$$\:\:\:\frac{{di}}{{dt}}=\frac{\left.\mathrm{10}\:{A}\right]/{second}...}{} \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 31/Jul/18
thank you sir ��