Question and Answers Forum

All Questions      Topic List

Magnetic Effect Questions

Previous in All Question      Next in All Question      

Previous in Magnetic Effect      Next in Magnetic Effect      

Question Number 41434 by rahul 19 last updated on 07/Aug/18

Commented by rahul 19 last updated on 07/Aug/18

$$\mathrm{If}\mathrm{the}\mathrm{ammeter}{\mathrm{A}}_2\mathrm{reads}1.6\mathrm{A}\mathrm{and}$$$$\mathrm{ammeter}{\mathrm{A}}_3\mathrm{reads}0.4\mathrm{A}.$$$$\mathrm{Find}\mathrm{reading}\mathrm{of}\mathrm{ammeter}{\mathrm{A}}_1?$$

Answered by ajfour last updated on 07/Aug/18

$$let{V}_{ac}=V_0\sin \omega t$$$$-L\frac{{di}_2}{dt}=△V_L=-V_{ac}\Rightarrow i_2=\frac{V_0}{\omega L}\cos \omega t$$$$(\int {di}_2=-\int \frac{V_{ac}}{L}dt=-\int \frac{V_0}{L}\sin \omega tdt)$$$$\frac{\int i_{3}dt}{C}=△V_C=-V_{ac}\Rightarrow i_3=-\omega {CV}_0\cos \omega t$$$$i_1=i_2+i_3$$$$=(\frac{V_0}{\omega L}-\omega {CV}_0)\cos \omega t$$$${\left(i_1\right)}_{rms}=\frac{\frac{V_0}{\omega L}-\omega {CV}_0}{\sqrt{2}}=\frac{{\left(i_2\right)}_{max}-{\left(i_3\right)}_{max}}{\sqrt{2}}$$$$$$$$=\frac{1.6\sqrt{2}-0.4\sqrt{2}}{\sqrt{2}}=1.2A.$$$$$$

Commented by rahul 19 last updated on 07/Aug/18

$$\mathrm{\Delta }{\mathrm{V}}_{\mathrm{L}}=\mathrm{\Delta }{\mathrm{V}}_{\mathrm{C}}=-{\mathrm{V}}_{\mathrm{AC}}?\mathrm{pls}\mathrm{explain}?$$

Commented by ajfour last updated on 07/Aug/18

$$KVL$$

Commented by ajfour last updated on 07/Aug/18

$$ammeterreadingsaretherms$$$$values{i}_{rms}=\frac{i_{max}}{\sqrt{2}}.$$

Commented by rahul 19 last updated on 07/Aug/18

Thank you sir!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com