91Ó°ÊÓ

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

$V=IR$

An inductor act as a wire of infinite resistance right after the circuit is closed while it acts as a normal conducting wire after a long time of circuit being closed

Same current flows through the ammeters connected in series.

Same voltage is dropped across voltmeters connected in parallel

Just after the switch is closed, the inductor has infinite resistance and hence the circuit branch is considered open at this point hence no current flows through the two ammeters$A_2$and $A_3$hence they read zero.

$A_2=A_3=0$

The same current flow through the ammeters $A_1\mathrm{and}{A}_4$since they are in series.

Use ohm’s law to get the current through them

$$\begin{gathered} i=\frac{\mathrm{Î\mu }}{40\mathrm{Î\copyright }+15\mathrm{Î\copyright }} \\ =\frac{25}{40\mathrm{Î\copyright }+15\mathrm{Î\copyright }} \\ =0.455A \end{gathered}$$

Hence the current through the ammeters is $A_1=A_4=0.455A$And$A_2=A_3=0$

After a long time the current in the circuit is in steady-state, voltage across the inductors is zero. We get current in each ammeter by ohm’s law

$$\begin{gathered} A_1=\frac{25V}{42.73\mathrm{Î\copyright }}=0.58A \\ {A}_2=\frac{1.6V}{5\mathrm{Î\copyright }}=0.32A \\ {A}_3=\frac{1.6V}{10\mathrm{Î\copyright }}=0.16A \\ {A}_4=\frac{1.6V}{15\mathrm{Î\copyright }}=0.107A \end{gathered}$$

Hence the current through the ammeters is $A_1=0.58A,A_2=0.32A,A_3=0.16A\mathrm{and}{A}_4=0.107A$,