In the circuit shown in Fig., E = 24.0 V,R₁ = 6.00$\Omega$,$R_3$= 12.0$\Omega$, and R₂ can vary between 3.00$\Omega$and 24.0$\Omega$. For what value ofR₂ is the power dissipated by heating elementR₁the greatest? Calculate the magnitude of the greatest power.

Consider the following circuit, where $\epsilon$= 24.0 v, R₁= 6.00$\Omega R_3$= 12.0 $\Omega$, and R₂can vary between 3.00$\Omega$and 24.0 $\Omega$. We need to find R₂such that the power dissipated by the heating element R₁is the greatest. According to the circuit, the power dissipated by the heating element R₁equals 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 I₁in terms of the resistor R₂. From the junction rule, we can find that,

I= I₁+ 1₂ (2)

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

$\epsilon -I_1{R}_1-{I_R}_3=0$ (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

$\epsilon -I_1{R}_1-I_2{R}_3\left(\frac{R_1+R_2}{R_2}\right)=0$

but $R_3=2R_1$, so we get

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

substituting into (1) we get,

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

or,

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

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

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

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 $24\Omega$, so,$R_3=24\Omega$

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

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