Question

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

Short answer

The potential at points $a\mathrm{and}b$are${\mathit{V}}_{\mathbf{a}}\mathbf{=}\mathbf{9}\mathbf{}\mathit{V}\mathbf{}\mathrm{and}\mathbf{}{\mathit{V}}_{\mathbf{b}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathit{V}$

Step-by-step solution

Given information and formula used 

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.

Finding the value of current

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 :

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

Finding value of the potential at points a and b

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 :

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