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An electronic circuit is made up of individual electronic components including resistors, transistors, capacitors, inductors, and diodes that are linked by conductive wires or traces.

Considering an electric circuit which has two loops and then calculating the potential differences at various points. To calculate the potential drops, we are suppose to calculate three currents which goes through the circuit. To do this, we first do the simplification of the circuit by adding all resistances together on the path of one current. We define the resistances as:

\(\begin{aligned}{}R{\rm{ }} & = {\rm{ }}{R_1} + {r_1} + {R_4}\\ & = {\rm{ }}(20 + 0.5 + 15){\rm{ }}\Omega \\ & = {\rm{ }}35.5{\rm{ }}\Omega \end{aligned}\)

\(\begin{aligned}{}{R'}{\rm{ }} & = {\rm{ }}{R_2} + {r_2}\\ & = {\rm{ }}(6 + 0.25){\rm{ }}\Omega \\ & = {\rm{ }}6.25{\rm{ }}\Omega \end{aligned}\)

\(\begin{aligned}{}{R^{''}}{\rm{ }} & = {\rm{ }}{R_3} + {r_3} + {r_4}\\ & = {\rm{ }}(8 + 0.5 + 0.75){\rm{ }}\Omega \\ & = {\rm{ }}9.25{\rm{ }}\Omega \end{aligned}\)

To obtain the currents, we use the Kirchhoff's rules. We then can use the junction rule for the junction\({\rm{b}}\), which then gives:

\({I_1}{\rm{ }} = {\rm{ }}{I_2} + {I_3}\)

The junction\({\rm{e}}\)must also have given us the same condition. The two other conditions (we need three for three currents) are the loop conditions for the loops\({\rm{abef}}\)and\({\rm{bcde}}\). We can also use the loop rule for the loop\({\rm{abcdef}}\), but it will not give us any new information. We therefore have:

\({E_1} - {I_1}R - {I_3}{R'} + {E_2}{\rm{ }} = {\rm{ }}0\)

\( - {I_2}{R^{''}} + {E_3} - {E_4} + {I_3}{R'} - {E_2}{\rm{ }} = {\rm{ }}0\)

The voltage across a resistor drops when it is moved in the direction of current. If it goes in the opposite direction, the voltage difference will be \( + IR\). When it is gone from the positive to the negative pole of the battery , the voltage difference will be negative.

Inserting the first condition into the second to get the third condition.

Then we obtain:

\(\begin{aligned}{}{E_1} - \left( {{I_2} + {I_3}} \right)R - {I_3}{R'} + {E_2}{\rm{ }} & = {\rm{ }}0\\ - {I_2}{R{''}} + {E_3} - {E_4} + {I_3}{R'} - {E_2}{\rm{ }} & = {\rm{ }}0\\{I_2}R + {I_3}\left( {R + {R'}} \right){\rm{ }} & = {\rm{ }}{E_1} + {E_2}\\{I_2}{R^{''}} - {I_3}{R'}{\rm{ }} & = {\rm{ }} - {E_2} + {E_3} - {E_4}\\{I_2}{\rm{ }} & = {\rm{ }} - {I_3}\frac{{R + {R'}}}{R} + \frac{{{E_1} + {E_2}}}{R}\\{I_2}{\rm{ }} & = {\rm{ }}{I_3}\frac{{{R'}}}{{{R{''}}}} - \frac{{{E_2} - {E_3} + {E_4}}}{{{R^{''}}}}\\ - {I_3}\frac{{R + {R'}}}{R} + \frac{{{E_1} + {E_2}}}{R}{\rm{ }} & = {\rm{ }} + {I_3}\frac{{{R'}}}{{{R{''}}}} - \frac{{{E_2} - {E_3} + {E_4}}}{{{R{''}}}}\\{I_3}\left( {\frac{{{R'}}}{{{R{''}}}} + \frac{{R + {R'}}}{R}} \right){\rm{ }} & = {\rm{ }}\frac{{{E_1} + {E_2}}}{R} + \frac{{{E_2} - {E_3} + {E_4}}}{{{R{''}}}}\end{aligned}\)

The currents are no being evaluated:

\(\begin{aligned}{}{I_3}{\rm{ }} & = {\rm{ }}\frac{{\left( {{E_1} + {E_2}} \right){R{''}} + \left( {{E_2} - {E_3} + {E_4}} \right)R}}{{R{R'} + R{R{''}} + {R'}{R{''}}}}\\ & = {\rm{ }}\frac{{((18 + 3)V) \times (9.25{\rm{ }}\Omega ) + ((3 - 12 + 24)V) \times (35.5{\rm{ }}\Omega )}}{{(35.5{\rm{ }}\Omega ) \times (6.25{\rm{ }}\Omega ) + (35.5{\rm{ }}\Omega ) \times (9.25{\rm{ }}\Omega ) + (6.25{\rm{ }}\Omega ) \times (9.25{\rm{ }}\Omega )}}\\ & = {\rm{ }}1.195\;{\rm{ }}A\end{aligned}\)

\(\begin{aligned}{}{I_2} & = {\rm{ }}{I_3}\frac{{{R'}}}{{{R{''}}}}{\rm{ }} - {\rm{ }}\frac{{{E_2} - {E_3} + {E_4}}}{{{R^{''}}}}\\ & = {\rm{ }}1.195{\rm{ }} \times {\rm{ }}\frac{{6.25{\rm{ }}\Omega }}{{9.25{\rm{ }}\Omega }} - \frac{{(3{\rm{ }} - {\rm{ }}12{\rm{ }} + {\rm{ }}24)V}}{{9.25{\rm{ }}\Omega }}\\ & = {\rm{ }} - 0.81\;A\end{aligned}\)

\(\begin{aligned}{}{I_1}{\rm{ }} & = {\rm{ }}{I_1}{\rm{ }} + {\rm{ }}{I_2}\\ & = {\rm{ }}(1.195{\rm{ }} - {\rm{ }}0.81)A\\ & = {\rm{ }}0.382\;A\end{aligned}\)

The fact that the sign of\({I_2}\)is a negative one then it means that we assumed the wrong direction of this current.

\(\begin{aligned}{}{V_{ab}}{\rm{ }} & = {\rm{ }} - {I_1}{R_1}\\ & = {\rm{ }} - (0.382{\rm{ }}A) \times (20{\rm{ }}\Omega )\\ & = {\rm{ }} - 7.64{\rm{ }}V\end{aligned}\)

\(\begin{aligned}{}{V_{cb}}{\rm{ }} & = {\rm{ }}{I_2}{R_3}\\& = {\rm{ }}( - 0.81{\rm{ }}A) \times (8{\rm{ }}\Omega )\\ & = {\rm{ }} - 6.48{\rm{ }}V\end{aligned}\)

\(\begin{aligned}{}{V_{eg}}{\rm{ }} & = {\rm{ }}{I_3}{r_2}{\rm{ }} - {\rm{ }}{E_3}\\ & = {\rm{ }}(1.95{\rm{ }}A) \times (0.25{\rm{ }}\Omega ){\rm{ }} - 3{\rm{ }}V\\ & = {\rm{ }} - 2.7{\rm{ }}V\end{aligned}\)

\(\begin{aligned}{}{V_{ed}}{\rm{ }} & = {\rm{ }}{I_2}{r_4}{\rm{ }} + {\rm{ }}{E_4}\\& = {\rm{ }}( - 0.81{\rm{ }}A) \times (0.75{\rm{ }}\Omega ){\rm{ }} + 24\\ & = {\rm{ }}23.39{\rm{ }}V\end{aligned}\)

Therefore, we get:

(a) Potential difference from point \(a\) to \(b\) is: \({V_{ab}}{\rm{ }} = {\rm{ }} - 7.64{\rm{ }}V\).

(b) Potential difference from point\(c\) to \(b\) is: \({V_{cb}}{\rm{ }} = {\rm{ }} - 6.48{\rm{ }}V\).

(c) Potential difference from point\(e\) to \(g\) is: \({V_{eg}}{\rm{ }} = {\rm{ }} - 2.7{\rm{ }}V\).

(d) Potential difference from point\(e\) to \(d\) is: \({V_{ed}}{\rm{ }} = {\rm{ }}23.39{\rm{ }}V\).