Question

Apply the loop rule to loop aedcba in Figure 21.25

Short answer

The loop of aedcba is \(\left( {6{\rm{ }}\Omega } \right){I_1} + \left( {3{\rm{ }}\Omega } \right){I_2} = 18V\)

Step-by-step solution

Definition of Kirchhoff's law

Kirchhoff's first law defines currents at circuit junctions. It states that the sum of currents flowing into and out of a junction in an electrical circuit equals the sum of currents flowing out of the junction.

Given and formula to be used

$$\begin{gathered} {\mathrm{E}}_1=18\mathrm{V},{\mathrm{R}}_1=6{\mathrm{\Omega r}}_1=0.5\mathrm{\Omega } \\ {\mathrm{E}}_2=45\mathrm{V},{\mathrm{R}}_2=2.5{\mathrm{\Omega r}}_2=0.5\mathrm{\Omega } \\ {\mathrm{R}}_3=1.5\mathrm{\Omega } \end{gathered}$$

Applying Kirchhoff's loop rule we obtain:

$-\left({\mathrm{l}}_2-{\mathrm{l}}_1\right)\left({\mathrm{R}}_1\right)+{\mathrm{l}}_2\left({\mathrm{r}}_1\right)+{\mathrm{E}}_1+{\mathrm{l}}_2\left({\mathrm{R}}_2\right)=0$

Calculation to the loop of aedcba

In this problem we consider the loop aedcba with an emf of \({E_1} = 18\;\;V\), corresponding internal resistance \({r_1} = 0.5\;\Omega \), and external resistances \({R_1} = 6\;\Omega \) and \({R_2} = 2.5\;\Omega \). We go around the loop in the counterclockwise direction, which is opposite to the direction of both currents \({I_1}\) and \({I_2}\). In this direction the voltage across the emf drops and we have

\(\begin{align}{}{I_1}{R_1} + {I_2}{r_1} - {E_1} + {I_2}{R_2} &= 0\\{E_1} - {I_1}{R_1} - {I_2}\left( {{r_1} + {R_2}} \right) &= 0\\\left( {6{\rm{ }}\Omega } \right){I_1} + \left( {3{\rm{ }}\Omega } \right){I_2} &= 18V\end{align}\)

Hence, the equation is obtained as \(\left( {6{\rm{ }}\Omega } \right){I_1} + \left( {3{\rm{ }}\Omega } \right){I_2} = 18V\).