In Figure,${\mathbf{R}}_1\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{V}$, ${\mathbf{R}}_2\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{V}$and the ideal battery has emf$\mathbf{\epsilon }\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}$.

(a) What is the size of current$\mathbf{}{\mathbf{i}}_1$?

(b) What is the direction (left or right) of current$\mathbf{}{\mathbf{i}}_1$?

(c) How much energy is dissipated by all four resistors in 1.00 min?

$$\begin{gathered} R_2=18.0\Omega \\ {R}_1=6.00\Omega \\ V=12.0\text{\hspace{0.17em}V} \end{gathered}$$

Time,$t=1.00\min =60.0\text{\hspace{0.17em}s}$

Use the concept of series as well as parallel resistances. Using that, we can find total resistance. Then, we have to use Ohm’s law$\mathbf{V}\mathbf{=}\mathbf{IR}$to find total current.

Write the formula for series and the parallel resistance:

$$\begin{gathered} R_{series}=R_1+R_2 \\ {R}_{parallel}=\frac{R_1{R}_2}{R_1+R_2} \end{gathered}$$

Write the formula for the electrical energy:

$E=i_1^2{R}_{1}t$

Here all $R_2$are connected in parallel. So, the equivalent of those is R .

$$\begin{gathered} \frac{1}{R}=\frac{1}{R_2}+\frac{1}{R_2}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{18}+\frac{1}{18}+\frac{1}{18} \\ R=6.00\Omega \end{gathered}$$

Now $R_1$and R are in series, so, equivalent of those is$R\text{'}$.

$$\begin{gathered} R^{\text{'}}=R_1+R \\ {R}^{\text{'}}=6.00+6.00 \\ {R}^{\text{'}}=12.0\Omega \end{gathered}$$

Now, according to Ohm’s law, total current l is as follow:

$$\begin{gathered} I=\frac{V}{R\text{'}} \\ I=\frac{12.0}{12.0} \\ I=1.00A \end{gathered}$$

Now,current$i_1$ , here I is the total current through$RandR\text{'}$and R is the combination of three parallel resistances$R_2.$So, the current through each$R_2.$is one third that of the total current.

So,

$\begin{array}{c}{i}_1=\frac{I}{3}\\ =\frac{1.00}{3}\\ =0.333\text{\hspace{0.17em}A}\end{array}$

Direction of current:

Consider the positive current direction from thepositive terminal to the negative terminal. So, the current l is clockwise means it is rightwards.

Energy dissipated:

$E=I^{2}R\text{'}t$
Substitute the values and solve for the energy dissipated as:

$$\begin{gathered} E={1.00}^2\times 12.0\times 60.0 \\ E=720\text{\hspace{0.17em}J} \end{gathered}$$