91影视

Consider the following circuit, where $蔚$= 24.0 v, R鈧�= 6.00$惟R_3$= 12.0 $惟$, and R鈧俢an vary between 3.00$惟$and 24.0 $惟$. We need to find R鈧俿uch that the power dissipated by the heating element R鈧乮s the greatest. According to the circuit, the power dissipated by the heating element R鈧乪quals the squared current flowing through it multiplied by its resistance, that is,

$P=I_2^1{R}_1$ (1)

our mission is to find the current I鈧乮n terms of the resistor R鈧�. From the junction rule, we can find that,

I= I鈧�+ 1鈧� (2)

apply the loop rule to the left loop (clockwise), and we get,

$蔚-I_1{R}_1-{I_R}_3=0$ (3)

and apply it to the right loop (also clockwise), we get,

$I_1{R}_1=I_2{R}_2$ (4)

we need to solve (2), (3) and (4) for $l_1$, substitute from (4) into (2) with $l_0$, so we get

$I=I_1+I_1\frac{R_1}{R_2}=I_1\left(1+\frac{R_1}{R_2}\right)=I_1\left(\frac{R_1+R_2}{R_2}\right)$

substitute into (3) to eliminate 1, so we get

$蔚鈭�I_1{R}_1鈭�I_2{R}_3\left(\frac{R_1+R_2}{R_2}\right)=0$

but $R_3=2R_1$, so we get

$$\begin{gathered} 蔚鈭�I_1{R}_1鈭�2I_1{R}_1\left(\frac{R_1+R_2}{R_2}\right)=0 \\ 蔚鈭�I_1{R}_1\left(1+2\frac{R_1+R_2}{R_2}\right)=0 \\ 蔚鈭�I_1{R}_1\left(\frac{2R_1+3R_2}{R_2}\right)=0 \\ {I}_1=\frac{\mathcal{E}{R}_2}{R_1\left(2R_1+3R_2\right)} \end{gathered}$$

substituting into (1) we get,

$P_1={\left(\frac{蔚R_2}{R_1\left(2R_1+3R_2\right)}\right)}^2{R}_1$ (5)

or,

$P_1=\frac{{蔚}^2{R}_2^2}{R_1\left(4R_1^2+12R_1{R}_2+9R_2^2\right)}$

dividing the denominator and the numerator by $R_2^2$, so we get,

$P_1=\frac{{蔚}^2}{R_1\left(4R_1^2/R_2^2+12R_1/R_2+9\right)}$

this is the maximum when the denominator is minimum, and the denominator is minimum when$R_2$is maximum. The maximum allowed value for$R_2$is $24惟$, so,$R_3=24惟$

now we need to find the magnitude of the greatest power, substitute into (5), we get,

$$\begin{gathered} P_1={\left(\frac{\left(24.0\mathrm{V}\right)\left(24.0\mathrm{惟}\right)}{\left(6.00\mathrm{惟}\right)\left(2\right(6.00\mathrm{惟}\left)\right)+3\left(24.0\mathrm{惟}\right)}\right)}^2\left(6.00\mathrm{惟}\right)=7.84\mathrm{W} \\ {\mathrm{P}}_1=7.84\mathrm{W} \end{gathered}$$