In the circuit shown, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts. The resistor that dissipates the most power is

In the given circuit, $R_2, R_3$ and $R_4$ are parallel to each other

${\mathrm{R}}_1,{\mathrm{R}}^1$ are in series.

${\mathrm{V}}_1=\frac{{\mathrm{ER}}_1}{{\mathrm{R}}_1+{\mathrm{R}}^1}=\frac{\left(3\right)50}{\left(\frac{450}{7}\right)}=\frac{7}{3}\mathrm{V}$

${\mathrm{V}}_2=\mathrm{E}-{\mathrm{V}}_1=3-\frac{7}{3}=\frac{2}{3}\mathrm{V}\text{, }$

$P_1=\frac{V_1^2}{R_1}=\frac{(7/3{)}^2}{50}=\frac{49}{450}\mathrm{W}\text{; }$

$P_2=\frac{V_2^2}{R_2}=\frac{(2/3{)}^2}{50}=\frac{4}{450}\mathrm{W}$

${\mathrm{P}}_3=\frac{{\mathrm{V}}_2^2}{{\mathrm{R}}_3}=\frac{(2/3{)}^2}{60}=\frac{4}{540}\mathrm{W}\text{; }$

${\mathrm{P}}_4=\frac{{\mathrm{V}}_2^2}{{\mathrm{R}}_4}=\frac{(2/3{)}^2}{30}=\frac{4}{270}\mathrm{W}$

Among all these, $P_1$ is maximum.