A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of

Q: Why is the terminal voltage less than the EMF of the battery?

When a battery is discharging (providing current to a circuit), the current must flow through the battery's own internal materials. These materials have resistance (internal resistance, $r$). Pushing current through this resistance causes a voltage drop ($ir$) inside the battery. Therefore, the voltage available at the terminals ($V$) is the total EMF ($E$) minus this internal drop: $V = E - ir$.

Q: Can I solve this by finding the current first?

Absolutely! You know the power of the external resistor is $P_R = V \times i$. So, $0.5 = 2.5 \times i$, which gives $i = 0.2\text{ A}$. Then, from $ir = 0.5$, you get $r = 2.5\text{ }\Omega$. Finally, $P_r = i^2 r = (0.2)^2 \times 2.5 = 0.10\text{ W}$. Both methods are perfectly valid, but the ratio method is often faster.

Q: Why is the current 'i' the same for both the internal and external resistance?

Because the internal resistance of the battery and the external resistor are connected in series. In a series circuit, there is only one path for the electrons to flow, so the current must be identical at every point in that loop.

Q: What exactly does 'power dissipated' mean in this context?

Power dissipation refers to the rate at which electrical energy is converted into another form of energy, usually heat. In the external resistor, it might be heat or light. In the internal resistance, it is purely heat, which is why batteries get warm when you use them heavily.

Q: Is the internal resistance a tiny physical resistor soldered inside the battery?

No, it's not a separate component. It represents the inherent electrical resistance of the chemicals, electrolytes, and conductive plates that make up the battery itself. We just model it as a separate resistor in series with an ideal voltage source to make the math work.

Question

A battery of $3.0	ext{ V}$ is connected to a resistor dissipating $0.5	ext{ W}$ of power. If the terminal voltage of the battery is $2.5	ext{ V}$, the power dissipated within the internal resistance is

Options:

(A) $0.072	ext{ W}$

(B) $0.125	ext{ W}$

(C) $0.50	ext{ W}$

(D) $0.10	ext{ W}$

Sigma Percentile · Accepted Answer

### The Setup: A Real-World Battery

Imagine you are holding a standard battery. While it might say $3.0	ext{ V}$ on the label, the reality of physics dictates that no battery is perfect. Every real battery has some internal resistance, denoted by $r$, which acts like a tiny, invisible resistor hidden inside the casing.

When we connect this battery to an external circuit—in this case, a resistor $R$ that is happily dissipating $0.5	ext{ W}$ of power—current begins to flow. This current must fight its way through both the external resistor and the internal resistance of the battery itself.

### Decoding the Terminal Voltage

The voltage you actually measure across the battery's terminals when it is delivering current is called the **terminal voltage** ($V$). Because some voltage is 'lost' pushing the current through the internal resistance, the terminal voltage is always less than the ideal EMF ($E$) of the battery during discharge.

The master equation governing this is:

$$V = E - ir$$

We are given that the EMF $E = 3.0	ext{ V}$ and the terminal voltage $V = 2.5	ext{ V}$. Let's substitute these values to find the voltage drop across the internal resistance:

$$2.5 = 3.0 - ir$$

$$ir = 0.5	ext{ V}$$

This tells us that $0.5	ext{ V}$ is being consumed just to push the current through the battery's own internal structure.

### The Power of Ratios

Now, let's look at the external circuit. The terminal voltage $V$ is exactly the voltage applied across the external resistor $R$. By Ohm's law, we know:

$$V = iR = 2.5	ext{ V}$$

Here is where we can use a brilliant mathematical shortcut. Instead of calculating the current $i$ explicitly, let's compare the voltage drops. We can divide the voltage drop across the internal resistance by the voltage drop across the external resistor:

$$\frac{ir}{iR} = \frac{0.5}{2.5}$$

The current $i$ beautifully cancels out, leaving us with the ratio of the resistances:

$$\frac{r}{R} = \frac{1}{5}$$

### The Final Reveal

We are asked to find the power dissipated within the internal resistance ($P_r$). The formula for electrical power is $P = i^2R$. Since the internal resistance $r$ and the external resistance $R$ are in series, the exact same current $i$ flows through both of them.

This means the ratio of the power they dissipate is directly proportional to the ratio of their resistances:

$$\frac{P_r}{P_R} = \frac{i^2r}{i^2R} = \frac{r}{R}$$

We already found that $\frac{r}{R} = \frac{1}{5}$. Therefore:

$$\frac{P_r}{P_R} = \frac{1}{5}$$

Now, we simply substitute the known power of the external resistor, $P_R = 0.5	ext{ W}$:

$$P_r = \frac{1}{5} 	imes 0.5	ext{ W}$$

$$P_r = 0.10	ext{ W}$$

And there we have it! The internal resistance is quietly dissipating $0.10	ext{ W}$ of power as heat. This elegant ratio method saves time and reduces the chance of calculation errors, a crucial skill for competitive exams.

Animated Solution for Physics - Current Electricity: A battery of $3.0 V$ is connected to a resistor dissipating $0.5 W$ of power. If the terminal voltage of the battery is $2.5 V$ , the power dissipated within the internal resistance is

Select Answer:

Visualized Solution

Circuit Analysis

Terminal Voltage Equation

Voltage Drop Across Internal Resistance

Voltage Across External Resistor

Ratio of Resistances

Power Dissipation Ratio

Final Calculation

The Way Forward

The Sigma Insight: Cells, EMF, and Internal Resistance

The Setup

Decoding the Terminal Voltage

The Power of Ratios

The Final Reveal

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