Question

An external resistor with resistance R is connected toa battery that has emf E and internal resistance r. Let P be theelectrical power output of the source. By conservation of energyP is equal to the power consumed by R. What is the value of P inthe limit that R is (a) very small; (b) very large? (c) Show that the
power output of the battery is a maximum when R=r. What isthis maximum P in terms of E and r? (d) A battery has E=64.0 Vand r=4.00Ω. What is the power output of this battery whenit is connected to a resistor R, for R=2.00Ω,R=4.00Ω, andR=6.00Ω? Are your results consistent with the general resultthat you derived in part (b)?

Short answer

1. Power limit is Zero
2. Power limit is$\frac{{\epsilon }^2}{R}$
3. Maximum power is$P_{max}=\frac{{\epsilon }^2}{4r}$
4. Powers are$P_1=227W,P_2=256W$ and$P_3=245W$ it is not consistent.

Step-by-step solution

Important Concepts

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.

$\mathbf{\sum }_{\mathbf{}}\mathit{V}\mathbf{=}\mathbf{0}$

Power when R is very small

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

$P=\epsilon I-I^{2}r$

The term$\epsilon I$ 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

$I=\frac{\epsilon }{R+r}$

Where R < < r we get

$I=\frac{\epsilon }{r}$

Input this into the power equation

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

The power consumed when R is small is zero

Power when R is very large

When R > > r then

$$\begin{gathered} I=\frac{\epsilon }{R+r} \\ I=\frac{\epsilon }{R} \end{gathered}$$

Power dissipated is given by

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

We get

$P=\frac{{\epsilon }^2}{R}$

Power when R = r

When$R=r$

$$\begin{gathered} I=\frac{\epsilon }{R+r} \\ I=\frac{\epsilon }{2r} \end{gathered}$$

Plug this into the power equation

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

We get

$P=\frac{{\epsilon }^2}{4r}$

For given values of emf and resistors

For a given emf $\epsilon =64.0V$and $r=4\Omega$.We need to find P when

$R_1=2\Omega ,R_2=\Omega$and$R_3=6\Omega$

Find the current using for every resistance,

$I=\frac{\epsilon }{R+r}$

Then input this into the power equation

$P=\epsilon I-I^{2}r$

We get

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