In a circuit there is a battery with internal resistance r and EMF E, which is connected to external load resistance R as shown. Find value of R so that maximum power dissipates across R.

The correct answer is R = r.

Step 1: Understand the Circuit Configuration

From the circuit diagram provided in the image, we can see a source with EMF $E$ and internal resistance $r$ connected in series with an external load resistor $R$.

Step 2: Express Current in the Circuit

The total equivalent resistance of the series circuit is $R_{\text{eq}}=R+r$.

Using Ohm's Law, the current $I$ flowing through the circuit is given by:

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

Step 3: Derive the Power Dissipated across External Load R

The power $P$ dissipated across the external load resistance $R$ is:

$P=I^{2}R={\left(\frac{E}{R+r}\right)}^{2}R=\frac{E^{2}R}{{(R+r)}^2}$

Step 4: Condition for Maximum Power Transfer

To maximize $P$ with respect to $R$, we set the derivative of $P$ with respect to $R$ to zero:

$\frac{dP}{dR}=0$

Differentiating $P$ using the quotient rule:

$\frac{dP}{dR}=E^2·\frac{{(R+r)}^2·1-R·2(R+r)}{{(R+r)}^4}=0$

Simplifying the numerator:

${(R+r)}^2-2R(R+r)=0$

$(R+r)\left[\right(R+r)-2R]=0$

Since $R+r\ne 0$, we have:

$r-R=0\Rightarrow R=r$

Conclusion:

According to the Maximum Power Transfer Theorem, maximum power is transferred to the external load when the external load resistance $R$ equals the internal resistance $r$ of the source.