91Ó°ÊÓ

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

$V=IR$

According to Kirchhoff’s Lawthe sum of the voltages around the closed loop is equal to null.

$\underset{\mathbf{}}{\mathbf{∑}}\mathit{V}\mathbf{=}\mathbf{0}$

WWe want to get the consumed powerPwhen Ris very small. In this case the current Iis leaving the source at the higher potential terminal and the energy is being delivered to the external circuit where the rate is given by

$P=\mathrm{Î\mu }I-I^{2}r$

The term$\mathrm{Î\mu }I$ is the rate at which work is done by the battery and the term$I^{2}r$ is the rate at which electrical energy is dissipated by the internal resistance of the battery. Where the current is given by

$I=\frac{\mathrm{Î\mu }}{R+r}$

Where R < < r we get

$I=\frac{\mathrm{Î\mu }}{r}$

Input this into the power equation

$P=\mathrm{Î\mu }\frac{\mathrm{Î\mu }}{r}-{\left(\frac{\mathrm{Î\mu }}{r}\right)}^{2}rP=0$

The power consumed when R is small is zero

When R > > r then

$$\begin{gathered} I=\frac{\mathrm{Î\mu }}{R+r} \\ I=\frac{\mathrm{Î\mu }}{R} \end{gathered}$$

Power dissipated is given by

$$\begin{gathered} P=I^{2}R \\ P={\left(\frac{\mathrm{Î\mu }}{R}\right)}^{2}R \end{gathered}$$

We get

$P=\frac{{\mathrm{Î\mu }}^2}{R}$

When$R=r$

$$\begin{gathered} I=\frac{\mathrm{Î\mu }}{R+r} \\ I=\frac{\mathrm{Î\mu }}{2r} \end{gathered}$$

Plug this into the power equation

$$\begin{gathered} P=\mathrm{Î\mu }I-I^{2}r \\ P=\mathrm{Î\mu }\frac{\mathrm{Î\mu }}{2r}-{\left(\frac{\mathrm{Î\mu }}{2r}\right)}^{2}r \end{gathered}$$

We get

$P=\frac{{\mathrm{Î\mu }}^2}{4r}$

For a given emf $\mathrm{Î\mu }=64.0V$and $r=4\mathrm{Î\copyright }$.We need to find P when

$R_1=2\mathrm{Î\copyright },R_2=\mathrm{Î\copyright }$and$R_3=6\mathrm{Î\copyright }$

Find the current using for every resistance,

$I=\frac{\mathrm{Î\mu }}{R+r}$

Then input this into the power equation

$P=\mathrm{Î\mu }I-I^{2}r$

We get

$P_1=227W,P_2=256W$and$P_3=245W$