$[15$ points$]$ A battery with potential difference $10$ V is connected to several resistors as shown

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S$/8$ a$)[8$ points$]$ Calculate the equivalent resistance of the circuit.

$>R_{45}(R_4$&$R_5)$

$L_{eq}^{\text{t}\text{o}\text{t}\text{a}\text{l}}$

$=R_3+R_{eq}^1$

$=2+0\mathrm{.}74$

$R_{45}=3+4=7$

$R_1{3}_1=\frac{R_1{R}_2}{R_1+R_2}$

$=\frac{(15)}{1+5}=0\mathrm{.}83$

$\frac{R_{12}}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{\text{:}[R_{eq}^1],[],[R_{eq}^1]}{R_{12}+R_{45}}=(R_{12}\frac{?}{?}$& $R_{45})=0\mathrm{.}742$

$2/2$

b$)[2$ pts$]$ Calculate the current through the hatterv

$I=\frac{V_{\text{b}\text{a}\text{t}\text{t}}}{R_{eq}}=\frac{10(V)}{2\mathrm{.}74}=$

$\frac{1}{5}$

c$)[5$ points$]$ Calculate the voltage difference across each resistor.