In FIGURE EX28.30, what is the value of the potential at points $a\mathrm{and}b$?

Given :

Theory used :

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero. It is also sometimes called Kirchhoff's voltage law or Kirchhoff's second law.

The ground wire has no effect on the circuit's behavior, but it does provide a location on the circuit with zero potential, which is the same as the potential of the earth.

As a result, the potential zero is at point 'c'.

Assume we proceed around a loop, measuring potential differences among circuit parts along the way, and when we get back to the starting point, the algebraic sum of these differences is zero.

$\Sigma V=0$

Let's design a clockwise loop and solve equation (28.2) for it, where the voltage across the resistors $R_1=1\Omega \mathrm{and}{R}_2=2\Omega$is negative because the current is flowing in the same direction.

The emf ${\epsilon }_1=9V$is is positive because the current flows from the negative to the positive terminal in the battery, whereas the emf ${\epsilon }_2=6V$is negative because the current flows from the positive to the negative terminal.

$$\begin{gathered} \Sigma V=0 \\ \Rightarrow {\epsilon }_1+∆V₁+∆V₂-{\epsilon }_2=0 \\ \Rightarrow {\epsilon }_1+I{R}_1+I{R}_2-{\epsilon }_2=0 \\ \Rightarrow I=\frac{{\epsilon }_1-{\epsilon }_2}{R_1+R_2} \end{gathered}$$

Put the values to get :

$$\begin{gathered} I=\frac{{\epsilon }_1-{\epsilon }_2}{R_1+R_2} \\ =\frac{9V-6V}{1\Omega +2\Omega } \\ =1A \end{gathered}$$

The current is positive, indicating that it is moving in a clockwise direction. Because the current should move from the greater potential to the lower potential while moving clockwise, it passes through point 'c' before point 'a', resulting in a negative differential between the two places.

$V_c-V_a=-9Volts$

We may get

role="math" localid="1649330083106" $V_a=V_c+9V=0+9V=\boxed{\mathbf{9}\mathit{V}}$by putting the value of $V_c$by zero.

To calculate the voltage across the resistance, we use Ohm's law.

$∆V_1=I{R}_1=(1A)\left(1\Omega \right)=1V$

Because the current goes from the greater potential to the lower potential when it passes through point 'b' before point 'c', the difference between the two points is positive.

$V_b-V_c=1V$

By plugging in the value of $V_c$by zero :

$V_b=V_c+1V=0+1V=\boxed{\mathbf{1}\mathit{V}}$