A cell of internal resistance r drives current through an external resistance R. The power delivered

Q: Why do we differentiate with respect to $R$ and not $r$?

Because the internal resistance $r$ is a fixed, inherent property of the given battery. We cannot change it. The external resistance $R$ is the variable load that we have the freedom to adjust to maximize the power.

Q: What happens to the power if $R$ is very small (close to zero)?

If $R$ is very small, the circuit is almost short-circuited. The current $I$ becomes very high, but since Power $P = I^2 R$, the multiplication by a near-zero $R$ causes the total power delivered to the external load to approach zero.

Q: What happens to the power if $R$ is very large (approaching infinity)?

If $R$ is very large, the circuit behaves almost like an open circuit. The current $I$ approaches zero. Even though $R$ is large, the $I^2$ term drops to zero much faster, causing the total power to approach zero.

Q: Is the efficiency of the circuit 100% when maximum power is transferred?

No, it is exactly 50%. When $R = r$, the power dissipated in the external load ($I^2 R$) is exactly equal to the power wasted as heat inside the battery ($I^2 r$). Half the energy is lost internally.

Q: Does this theorem apply to AC circuits as well?

Yes, but with a slight modification. In AC circuits, to achieve maximum power transfer, the load impedance must be the complex conjugate of the source impedance. This means the resistances must be equal, and the reactances must be equal in magnitude but opposite in sign to cancel each other out.

Question

A cell of internal resistance $r$ drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when

Options:

(A) $R = 2r$

(B) $R = r$

(C) $R = 0.001 r$

(D) $R = 1000 r$

Sigma Percentile · Accepted Answer

## The Maximum Power Transfer Theorem

Imagine you are designing a circuit to extract the absolute maximum amount of power from a battery to run a motor or light up a bulb. You might intuitively think, "Let's just use the smallest possible resistance so a massive current flows!" Or perhaps, "Let's use a huge resistance so the voltage drop is maximum!"

As it turns out, nature has a much more elegant, balanced solution. Let's dive into the beautiful mathematics behind the **Maximum Power Transfer Theorem**.

### Analyzing the Setup

Every real-world power source, whether it's a tiny AA battery or a massive generator, has some internal friction. We model this as an ideal voltage source (EMF, denoted by $E$) in series with an internal resistance ($r$).

When we connect an external load resistance ($R$) to this battery, a current $I$ begins to flow. According to Ohm's Law, the total current in the circuit is simply the total voltage divided by the total resistance:

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

### The Master Equation for Power

The power $P$ that is actually delivered to our useful external load $R$ is given by Joule's heating law:

$$P = I^2 R$$

Substituting our expression for the current $I$ into this power equation, we get our master equation:

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

Let's pause and look at this equation. 
If $R$ is extremely small (close to $0$), the numerator becomes tiny, and the power $P$ approaches $0$. 
Conversely, if $R$ is extremely large (approaching infinity), the $R^2$ term in the denominator dominates, and the power $P$ again approaches $0$.

Since the power is zero at both extremes and positive in between, there *must* be a sweet spot—a specific value of $R$ where the power hits an absolute maximum!

### Calculus to the Rescue

To find this exact peak, we call upon the power of calculus. We need to differentiate the power $P$ with respect to the variable load $R$ and set the derivative to zero:

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

Applying the quotient rule to our master equation:

$$\frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2 \cdot (1) - R \cdot 2(R+r)}{(R+r)^4} \right] = 0$$

For this massive fraction to equal zero, its numerator must be exactly zero. Let's extract the numerator and simplify it:

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

We can factor out a common $(R+r)$ term:

$$(R+r)(R+r - 2R) = 0$$

$$(R+r)(r - R) = 0$$

### The Beautiful Conclusion

Since resistance cannot be negative, $R+r$ cannot be zero. Therefore, the only way this equation holds true is if:

$$r - R = 0 \implies \mathbf{R = r}$$

This is the legendary **Maximum Power Transfer Theorem**. It states that to extract the maximum possible power from a source, the resistance of your external load must perfectly match the internal resistance of the source.

**A Fascinating Caveat:** 
While $R=r$ gives you maximum power, it does *not* give you maximum efficiency. When $R=r$, exactly half of the total power generated is dissipated as useless heat inside the battery's internal resistance $r$. Therefore, at maximum power transfer, the efficiency of the circuit is exactly **50%**!

Animated Solution for Physics - Current Electricity: A cell of internal resistance $r$ drives current through an external resistance $R$ . The power delivered by the cell to the external resistance will be maximum when

Select Answer:

Visualized Solution

Circuit Setup

Current in the Circuit

Power Formula

Master Equation for Power

Condition for Maximum Power

Applying Quotient Rule

Solving the Equation

Maximum Power Transfer Theorem

Efficiency at Maximum Power

The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup

The Master Equation for Power

Calculus to the Rescue

The Beautiful Conclusion

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