91Ó°ÊÓ

Power is defined as the energy per unit of time and can be expressed as,

$\mathbf{P}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{R}\mathbf{ }\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

$$\begin{gathered} \mathrm{Î\mu }=45 V \\ {R}_1=5.00 \mathrm{Î\copyright } \\ {R}_2=10.0 \mathrm{Î\copyright } \\ {R}_3=15.0 \mathrm{Î\copyright } \\ {R}_4=20.0 \mathrm{Î\copyright } \end{gathered}$$

The equivalent resistance of three parallel resistors,

$$\begin{gathered} \frac{1}{{\mathrm{R}}^{\mathrm{′}}}=\frac{1}{{\mathrm{R}}_4}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_4} \\ \frac{1}{{\mathrm{R}}^{\mathrm{′}}}=\frac{1}{{\mathrm{R}}_4}+\frac{2}{{\mathrm{R}}_4}+\frac{1}{{\mathrm{R}}_4} \\ \frac{1}{{\mathrm{R}}^{\mathrm{′}}}=\frac{4}{{\mathrm{R}}_4} \\ {\mathrm{R}}^{\mathrm{′}}=\frac{{\mathrm{R}}_4}{4} \end{gathered}$$

Current is the summation of EMFs divided by the summation of resistance,

$$\begin{gathered} I=\frac{\mathrm{Î\mu }}{\left(R_1+R_2+R_3+\frac{R_4}{4}\right)} \\ I=\frac{45\mathrm{V}}{5.00\mathrm{Î\copyright }+10.00\mathrm{Î\copyright }+15\mathrm{Î\copyright }+5.0\mathrm{Î\copyright }} \\ =1.28\mathrm{A} \end{gathered}$$

Use junction rule,

$I=I_1+I_2+I_3$ (2)

V=IR thus,

$$\begin{gathered} I_1{R}_4=I_2{R}_2=I_3{R}_4 \\ 2I_1{R}_2=I_2{R}_2=2I_3{R}_2 \\ 2I_1=I_2=2I_3 \end{gathered}$$

Substituting the values in equation (2), we get

$$\begin{gathered} I=I_1+2I_1+I_1 \\ I=4I_1 \\ {I}_1=\frac{1.28A}{4} \\ {I}_1=0.32A \end{gathered}$$

The rate of energy in the resistor is,

$$\begin{gathered} P=I^2{R}_4 \\ =\left(0.322\mathrm{A}{)}^2\right(20\mathrm{Î\copyright }) \\ =2.07\mathrm{W} \end{gathered}$$

is the heat required to melt the ice of mass dm, and L is the latent heat of fusion. thus,

$$\begin{gathered} \mathrm{dE}=\mathrm{Ldm} \\ \frac{\mathrm{dE}}{\mathrm{dt}}=\mathrm{L}\frac{\mathrm{dm}}{\mathrm{dt}} \\ \frac{\mathrm{dm}}{\mathrm{dt}}=\frac{\mathrm{P}}{\mathrm{L}} \\ \frac{\mathrm{dm}}{\mathrm{dt}}=\frac{2.07\mathrm{W}}{3.34×{10}^5\mathrm{J}/\mathrm{kg}} \\ \frac{\mathrm{dm}}{\mathrm{dt}}=6.21×{10}^{−6}\mathrm{kg}/\mathrm{s}\left(\frac{1000\mathrm{g}/\mathrm{s}}{1\mathrm{kg}/\mathrm{s}}\right) \\ \frac{\mathrm{dm}}{\mathrm{dt}}=6.21×{10}^{−3}\mathrm{g}/\mathrm{s} \end{gathered}$$

Thus, the circuit will melt ice at $\frac{dm}{dt}=6.21×{10}^{-3}g/s$.