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

The emf of the battery is

(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.

$$\begin{gathered} V=I\left(R_2+R_3\right)\backslash \mathrm{hfill} \\ I=\frac{V}{R_2+R_3} \end{gathered}$$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(1)

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,

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

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,

$$\begin{gathered} V_{ab}=I{R}_2 \\ =\left(4.0A\right)×\left(6.0\mathrm{Î\copyright }\right) \\ =24.0V \end{gathered}$$

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,

$$\begin{gathered} R_{12}=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{4\mathrm{Î\copyright }×6\mathrm{Î\copyright }}{4\mathrm{Î\copyright }+6\mathrm{Î\copyright }} \\ =2.4\mathrm{Î\copyright } \end{gathered}$$

The current in the circuit will change, and the new current () could be calculated using equation (1) in the next form

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

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,

$$\begin{gathered} V_{ab}=I_{12}{R}_{12} \\ {V}_{ab}=\left(6.67A\right)×\left(2.4\mathrm{Î\copyright }\right) \\ {V}_{ab}=16V \end{gathered}$$

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$I_2=I_3=4A$

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

$I_1=0A$

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

$$\begin{gathered} I_1=\frac{V_{ab}}{R_1} \\ {I}_1=\frac{16V}{4\mathrm{Î\copyright }} \\ {I}_1=4A \\ {I}_2=\frac{V_{ab}}{R_2} \\ {I}_2=\frac{16V}{6\mathrm{Î\copyright }} \\ {I}_2=2.67A \end{gathered}$$

Therefore when S is openandand when S is closedincreasesandincreases

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