A triangular array of resistors is shown in Fig. E. What current will this array draw from a 35.0-V battery having negligible internal resistance if we connect it across

(a) ab ;

(b) bc;

(c) ac ;

(d) If the battery has an internal resistance of $\mathbf{3}\mathbf{.}\mathbf{00}\mathit{\Omega }$, what current will the array draw if the battery is connected across bc?

The equivalent resistance is where the resistance connected either in parallel or series is calculated.

Let $R_1=15\hspace{0.33em}\Omega ,R_2=10\hspace{0.33em}\Omega$, and $R_3=20\hspace{0.33em}\Omega$. The voltage of the battery is V =35.0 V.

a)

When the battery is connected across ab, the two resistors${\mathit{R}}_{\mathbf{2}}$and ${\mathit{R}}_{\mathbf{3}}$will be in series as shown in figure 1 and their combination$R_{23}=R_2+R_3=30\hspace{0.33em}\Omega$is in parallel with${\mathit{R}}_1$.

Where the current through resistors connected in parallel is the sum of the currents through the individual resistors${\mathit{R}}_1$and${\mathit{R}}_{\mathbf{23}}$, so the current acrosswill be given by Ohm’s law.

$I_t=I_{ab}=\frac{V}{R_{eq}}$ (1)

Where${\mathit{R}}_{\mathbf{e}\mathbf{q}}$represents the equivalent resistance in the circuit and it can be given as:

$$\begin{gathered} R_{\text{eq}}=\frac{R_1{R}_{23}}{R_1+R_{23}} \\ =\frac{\left(15\right)\left(30\right)}{15+30} \\ =10\mathrm{\Omega } \end{gathered}$$

Plug these values into equation (1),

$$\begin{gathered} l_{ab}=\frac{V}{R_{eq}} \\ =\frac{35.0\mathrm{V}}{10\mathrm{\Omega }} \\ =3.5\mathrm{A} \end{gathered}$$

The current flowing through ab is 3.5 A.

b)

As shown in figure 2, when the battery is connected across bc, the two resistors ${\mathit{R}}_{\mathbf{1}}$and ${\mathit{R}}_3$will be in series and their combination $R_{13}=R_1+R_3=35\hspace{0.33em}\Omega$is in parallel with ${\mathit{R}}_2$.

With the same steps as in part (a), the equivalent resistance can be given as:

$$\begin{gathered} R_{\text{eq}}=\frac{R_2{R}_{13}}{R_2+R_{13}} \\ =\frac{\left(10\right)\left(35\right)}{10+35} \\ =7.78\mathrm{\Omega } \end{gathered}$$

Plug these values in equation (1),

$$\begin{gathered} I_{bc}=\frac{V}{R_{\text{eq}}} \\ =\frac{35.0\mathrm{V}}{7.78\mathrm{\Omega }} \\ =4.5\mathrm{A} \end{gathered}$$

The current flowing through bc is 4.5 A.

c)

With the same steps as above, the equivalent resistance is given as:

$$\begin{gathered} R_{\text{eq}}=\frac{R_3{R}_{12}}{R_3+R_{12}} \\ =\frac{\left(20\right)\left(25\right)}{20+25} \\ =11.11\mathrm{\Omega } \end{gathered}$$

Plug these values in equation (1),

$$\begin{gathered} I_{ac}=\frac{V}{R_{eq}} \\ =\frac{35.0\mathrm{V}}{11.11\mathrm{\Omega }} \\ =3.15\mathrm{A} \end{gathered}$$

The current flowing through ac is 3.15 A.

d)

Here, $\mathit{r}\mathbf{=}\mathbf{3}\mathbf{\hspace{0.33em}}\mathit{\Omega }$

In part (b), the voltage drop is only due to ${\mathit{R}}_{\mathbf{e}\mathbf{q}}$but here it is due to ${\mathit{R}}_{\mathbf{eq}}$and r, so the current across bc is related to the emf of the battery, and its internal resistance is given as:

${\mathit{I}}_{\mathbf{b}\mathbf{c}}\mathbf{=}\frac{\mathbf{\epsilon }}{{\mathbf{R}}_{\mathbf{e}\mathbf{q}}\mathbf{+}\mathbf{r}}$ (2)

Where ${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{78}\mathbf{\hspace{0.33em}}\mathit{\Omega }$, Now, plug these values in equation (2),

$\begin{array}{r}{I}_{bc}=\frac{35.0\mathrm{V}}{7.78\mathrm{\Omega }+3\mathrm{\Omega }}\\ =3.25\mathrm{A}\end{array}$

The current decreases due to the internal resistance of the battery.

Thus, the current through bc is 3.25 A.