Physics - EMI/AC Question with Solution | TestHub

$$\begin{array}{rl}& {\mathrm{I}}_1=\frac{\mathrm{F}}{{\mathrm{R}}_1}=\frac{12}{2}=6\mathrm{A}\\ & \mathrm{E}=\mathrm{L}\frac{\mathrm{d}{\mathrm{l}}_2}{\mathrm{dt}}+{\mathrm{R}}_2\times {\mathrm{I}}_2\\ & {\mathrm{I}}_2={\mathrm{I}}_0\left(1-{\mathrm{e}}^{-\mathrm{t}/{\mathrm{t}}_{\mathrm{c}}}\right)\Rightarrow {\mathrm{I}}_{\mathrm{o}}=\frac{\mathrm{E}}{{\mathrm{R}}_2}=\frac{12}{2}=6\mathrm{A}\\ & {\mathrm{t}}_{\mathrm{c}}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{400\times 10^{-3}}{2}=0.2\\ & {\mathrm{I}}_2=6\left(1-{\mathrm{e}}^{-\mathrm{t}/0.2}\right)\end{array}$$ Potential drop across $L=E-R_2{l}_2=12-2\times 6\left(1-e^{-bt}\right)=12e^{-5t}$
Directions: Question numbers 28,29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle $ABCDA$, as shown in the $P-T$ diagram.