Question

In physics, Ohm's law says that current through a wire, I, is directly proportional to voltage, V, and inversely proportional to resistance, R:

I=V/R.
It's also true that resistance is directly proportional to the length of the wire. We have a piece of wire. We pass 12 volts through this wire and measure 100 milliamps of current. If I cut the wire in half and pass 24 volts through it, how many milliamps of current will I measure?

Answer 1

just a quick addition, to the risk of sounding redundant.

`\bf \qquad \qquad \textit{compound proportional variation} \\\\ \begin{array}{llll} \textit{\underline{y} varies directly with \underline{x}}\\ \textit{and inversely with \underline{z}} \end{array}\implies y=\cfrac{kx}{z}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] ~\dotfill`

`\bf \stackrel{\textit{I varies directly with V and inversely with R}}{I=\cfrac{(1)V}{R} }~\hfill \leftarrow k=1\to ~\hfill \stackrel{\textit{R varies directly with L}}{R=(1)L} \\\\[-0.35em] ~\dotfill\\\\ \textit{we know that } \begin{cases} V=12\\ I=100 \end{cases}\implies 100=\cfrac{(1)12}{R}\implies R=\cfrac{12}{100}\implies R=\cfrac{3}{25}`

now, we know R = L, so if we cut the length in half, namely L/2, will also mean we're cutting the resistance amount by half, R/2 = L/2 on that direct variation.

so what would the current "I" be if 3/25 gets halved and V = 24 volts ?

`\bf I=\cfrac{V}{R}\implies I=\cfrac{\stackrel{V}{24}}{\stackrel{\textit{halved R}}{(3)/(25)/ 2}}\implies I=\cfrac{24}{(3)/(25)\cdot (1)/(2)}\implies I=\cfrac{24}{(3)/(50)} \\\\\\ I=\cfrac{\stackrel{8}{~~\begin{matrix} 24 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{1}\cdot \cfrac{50}{~~\begin{matrix} 3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies I=400`