91Ó°ÊÓ

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

We are given${\mathrm{R}}_1=4.0\mathrm{Î\copyright },{\mathrm{R}}_2=6.0\mathrm{Î\copyright },{\mathrm{R}}_3=3.0\mathrm{Î\copyright }$

The emf of the battery is E=36. V

(a)

We Want V, when S is open and when it is closed.

When S is open, the current flow only in R and R, and the voltage drop of the battery is due to R₂ and R₃ As both these resistors are in series, therefore, they have the same current and We could get the current by using Ohm‘s law in the next form.

Now let us plug our values for V, R2, and R3 into equation (1) to get I which will help us to find the required voltage,

role="math" localid="1655716064912" $$\begin{gathered} \mathrm{I}=\frac{36\mathrm{V}}{6\mathrm{Î\copyright }+3\mathrm{Î\copyright }} \\ =4\mathrm{A} \end{gathered}$$

The voltage across ab is due to the voltage of R₂ so this voltage could be calculated by using Ohm's law in the next form,

role="math" localid="1655715865929" $$\begin{gathered} {\mathrm{V}}_{\mathrm{ab}}={\mathrm{IR}}_2 \\ =(4.0\mathrm{A})×(6.0\mathrm{Î\copyright }) \\ =24.0\mathrm{V} \end{gathered}$$

When S has closed the current flow in all resistors where R1 and R2 are in parallel and their combination could be

Therefore the potential difference between points a and b when S is open is 24.0 V

Calculated the equivalent resistance when R₁ and R₂ are in parallel combination,

$$\begin{gathered} {\mathrm{R}}_{12}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2} \\ =\frac{4\mathrm{Î\copyright }×6\mathrm{Î\copyright }}{4\mathrm{Î\copyright }+6\mathrm{Î\copyright }} \\ =2.4\mathrm{Î\copyright } \end{gathered}$$

The current in the circuit will change, and the new current $({\mathrm{I}}_2={\mathrm{I}}_3)$could be calculated using equation (1) in the next form

$$\begin{gathered} {\mathrm{I}}_{12}=\frac{36\mathrm{V}}{2.4\mathrm{Î\copyright }+3\mathrm{Î\copyright }} \\ =6.67\mathrm{A} \end{gathered}$$

As R₁ and R₂ are in parallel, therefore, the voltage across ab represents the voltage across their combination and it could be calculated by Ohm's law in the next form,

Therefore the potential difference between points a and b when S is closed is 16 V

(b)

Only in R2 and R3 and as both resistors are in series, therefore, the current is the same for both of them Where from part (a) the current is ${\mathrm{I}}_2={\mathrm{I}}_3=4\mathrm{A}$

When S is closed, the current flows in R₃ are the same as in part (a)

${\mathrm{I}}_1=0\mathrm{A}$

But for the combination R_m the current is cut into both resistors.

Where We can use the voltage across ab for each resistor.

To obtain the current in R₁ and R₂ using Ohm's law

$$\begin{gathered} {\mathrm{I}}_1=\frac{{\mathrm{V}}_{\mathrm{ab}}}{{\mathrm{R}}_1} \\ {\mathrm{I}}_1=\frac{16\mathrm{V}}{4\mathrm{Î\copyright }} \\ {\mathrm{I}}_1=4\mathrm{A} \\ {\mathrm{I}}_2=\frac{{\mathrm{V}}_{\mathrm{ab}}}{{\mathrm{R}}_2} \\ {\mathrm{I}}_2=\frac{16\mathrm{V}}{6\mathrm{Î\copyright }} \\ {\mathrm{I}}_2=2.67\mathrm{A} \end{gathered}$$

Therefore when S is open ${\mathrm{I}}_1=0\mathrm{A}$and ${\mathrm{I}}_2={\mathrm{I}}_3=4\mathrm{A}$and when S is closed ${\mathrm{I}}_1=4\mathrm{A}$increases ${\mathrm{I}}_2=2.67\mathrm{A}\mathrm{and}{\mathrm{I}}_3=6.67\mathrm{A}$ increases

The current through R₁ and R₃ increases while through R₂ it decreases.