(a) Find the potential of point a with respect to point b in Fig. P26.57. (b) If points a and b are connected by a wire with negligible resistance, find the current in the \({\bf{12}}.{\bf{0V}}\)battery.

Battery = \({\rm{12V}}\)

Resistance in \({\rm{12V,10V,and 8V }}\)= \(1.00\Omega \)

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The vertical resistance =

\(\begin{aligned}1.00\Omega {\rm{ and }}2.00\Omega \\2.00\Omega {\rm{ and }}2.00\Omega \end{aligned}\)

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The passive two terminal electrical component by which electrical resistance get implemented as a circuit element is known as resistor R.

\(\begin{aligned}{R_{Total}} = {R_1} + {R_2} + {R_3} + ...(series)\\{R_{Total}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + ...(parallel)\\R = \frac{V}{I}\\V = voltage\\I = current\end{aligned}\)

In the loop 1st the current equation according to the Kirchhoff’s law,

\(\begin{aligned}1{\rm{2V}} - I\left( {1.00\Omega + 2.00\Omega + 2.00\Omega + 1.00\Omega } \right) - 8.{\rm{0V}} - I\left( {2.00\Omega + 1.00\Omega } \right) = 0\\I = \frac{{12.0V - 8.0V}}{{9.00\Omega }}\\I = 0.4444{\rm{A}}\end{aligned}\)

Now to calculate the voltage between a add b we have to start from point b to a that is loop 2, so now,

\(\begin{aligned}{V_b} - 10{\rm{V + 12V}} - 0.4444{\rm{A}}\left( {1.00\Omega + 1.00\Omega + 2.00\Omega } \right) = {V_a}\\{V_a} - {V_b} = + 0.22{\rm{V}}\end{aligned}\)

It is showing potential a is at higher potential,

Hence, the potential of point a with respect to point b is \({V_a} - {V_b} = + 0.22{\rm{V}}\).