Consider the circuit shown in Fig.P26.63. (a) What must the emf Ԑ of the battery be for a current of 2.00 A to flow through the 5.00-V battery as shown? Is the polarity of the battery correct as shown? (b) How long does it take for 60.0 J of thermal energy to be produced in the$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{\Omega }$ resistor?

Kirchoff’s law state that at any junction the algebraic sum of all currents should be always equal to zero.

The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero

(a)

Consider the following circuit, where$R_1=5.00\hspace{0.33em}\Omega ,\hspace{0.33em}{R}_2=10.0\hspace{0.33em}\Omega ,\hspace{0.33em}{R}_3=15.0\hspace{0.33em}\Omega ,\hspace{0.33em}{R}_4=20.0\hspace{0.33em}\Omega ,$

$R_5=30.0\hspace{0.33em}\Omega ,\hspace{0.33em}{R}_6=60.0\hspace{0.33em}\Omega ,\hspace{0.33em}{\epsilon }_1=10.0\hspace{0.33em}V$

we need to find the emfand such that the current flowing through the 5.00-V battery is I = 2.00 A.

We can simplify the circuit by finding the equivalent resistance and the equivalent emfas shown in the second figure.

We know that the resistance adds in series and its reciprocal add in parallel, so we have,

$R\text{'}=2R_1+R_{parallel}$ (1)

where,

$\frac{1}{{\mathrm{R}}_{\mathrm{parallel}}}=\frac{1}{{\mathrm{R}}_1}+\frac{2}{{\mathrm{R}}_6}+\frac{1}{{\mathrm{R}}_2+{\mathrm{R}}_4}$

we can express all the resistance in terms of $R_1$, that is $R_2=R_1$, $R_3=3R_1,\hspace{0.33em}{R}_4=4R_1,\hspace{0.33em}{R}_5=6R_1,\hspace{0.33em}{R}_6=12R_1$and thus

$$\begin{gathered} \frac{1}{R_{\text{parallel}}}=\frac{1}{6R_1}+\frac{2}{12R_1}+\frac{1}{6R_1}=\frac{1}{2R_1} \\ {R}_{\text{parallel}}=2R_1 \end{gathered}$$

substitute into(1) we get,

$$\begin{gathered} R^{\mathrm{\prime }}=2R_1+2R_1 \\ =4R_1 \\ =20.0\mathrm{\Omega } \end{gathered}$$ (2)

also, we have,

$$\begin{gathered} \epsilon \text{'}={\epsilon }_2-{\epsilon }_1 \\ =5.00V \end{gathered}$$ (3)

assume that the current flowing through the resistor $R_3$in the second figure is $l_1$, and the current flowing in the resistor $R_4$, is $l_2$, from the junction rule we can write,

$I=I_1+I_2$ (4)

Now we can apply the loop rule two times first by passing through the branch where the resistor $R_3$exists and second by passing through the branch where the resistor localid="1668334078768" $R_4$exists (counterclockwise), so we get

${\epsilon }^{\text{'}}+I{R}^{\text{'}}+I_1{R}_3+\epsilon =0$ (5)

and,

${\epsilon }^{\text{'}}+I{R}^{\text{'}}+I_2{R}_4=0$ (6)

now we have three equations(4), (5), and (6), we need to solve them for $\epsilon$, first, subtract (6) from (5), we get,

$I_1{R}_3+\epsilon =I_2{R}_4$

or,

$l_2=\frac{l_1{R}_3}{R_4}+\frac{\epsilon }{R_4}$

but ${\mathrm{R}}_3=3{\mathrm{R}}_1,\hspace{0.33em}\mathrm{and}\hspace{0.33em}{\mathrm{R}}_4=4{\mathrm{R}}_1\hspace{0.33em}$, therefore,

${\mathrm{I}}_2=\frac{3{\mathrm{I}}_1}{4}+\frac{\epsilon }{4{\mathrm{R}}_1}$

substituting into (4) we get,

$\begin{array}{r}\mathrm{I}=\frac{7{\mathrm{I}}_1}{4}+\frac{\epsilon }{4{\mathrm{R}}_1}\\ {\mathrm{I}}_1=\frac{4}{7}\mathrm{I}+\frac{\epsilon }{7{\mathrm{R}}_1}\end{array}$

substitute (5) to get,

$$\begin{gathered} {\epsilon }^{\mathrm{\prime }}+I{R}^{\mathrm{\prime }}+\frac{4}{7}I{R}_3-\frac{\epsilon R_3}{7R_1}+\epsilon =0 \\ {\epsilon }^{\mathrm{\prime }}+I{R}^{\mathrm{\prime }}+\frac{12}{7}I{R}_1-\frac{3\epsilon }{7}+\epsilon =0 \\ \epsilon =-\frac{4}{7}\left({\epsilon }^{\mathrm{\prime }}+I{R}^{\mathrm{\prime }}+\frac{12}{7}\mid R_1\right) \end{gathered}$$

substitute with the given to get,

$$\begin{gathered} \epsilon =-\frac{7}{10}\left(5.00\mathrm{V}+\left(2.00\mathrm{A}\right)\left(20.0\mathrm{\Omega }\right)+\frac{12}{7}\left(2.00\mathrm{\Omega }\right(5.00\mathrm{\Omega }\left)\right)\right) \\ \epsilon =-109\mathrm{V} \end{gathered}$$

Since the sign of the emf is negative, then the polarity of the battery is not correct and it must be reversed

(b) Now we need to find how long it takes for 6.00 J of thermal energy to be produced in the $10.0\hspace{0.33em}\Omega$resistor, to do that we need to find the current flowing through this resistor, the current flowing through the parallel resistors network is I= 2.00 A, and the potential across each one of the resistors is the same, which equals the current flowing the parallel resistors network multiplied by their resistance, $R_{parallel}$that is

$$\begin{gathered} V=I{R}_{parallel} \\ =2I{R}_1 \end{gathered}$$

the current flowing through the resistor R₂equals this potential divided by the resistance of its branch, that is.

$$\begin{gathered} {\mathrm{I}}_2=\frac{\mathrm{V}}{{\mathrm{R}}_2+{\mathrm{R}}_4} \\ =\frac{2\mid {\mathrm{R}}_1}{{\mathrm{R}}_2+{\mathrm{R}}_4} \\ =\frac{2\mid {\mathrm{R}}_1}{2{\mathrm{R}}_1+4{\mathrm{R}}_1} \\ =\frac{1}{3} \end{gathered}$$

the energy of this resistor equals its power multiplied by the time, that is

$$\begin{gathered} {\epsilon }_2=I_2^2{R}_{2}t \\ =\frac{l^2{R}_{1}t}{9} \end{gathered}$$

solve for t to get

$t=\frac{9{\epsilon }_2}{2l^2{R}_1}$

substitute with $I=2.00A,\hspace{0.33em}{R}_1=5.00\hspace{0.33em}\Omega ,$and ${\epsilon }_2=60.0\hspace{0.33em}J$

$$\begin{gathered} t=\frac{9\left(60.0\mathrm{J}\right)}{2\left(2.00\mathrm{A}{)}^2\right(5.00\mathrm{\Omega })} \\ t=13.5\mathrm{s} \end{gathered}$$