To verify Ohm's law, a student connects the voltmeter across the battery as shown in the figure. The

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

When current flows through the circuit, the battery's internal resistance causes a small voltage drop ($Ir$) inside the battery itself. The terminal voltage is what's left over: $V = E - Ir$.

Q: What does the slope of the V-I graph represent?

If you rearrange the equation to $V = -rI + E$, it matches the equation of a line $y = mx + c$. The slope $m$ is exactly $-r$, the negative of the internal resistance.

Q: Why did we convert 1000 mA to 1 A?

To keep our units consistent. The EMF is in Volts and resistance is in Ohms. To use Ohm's law correctly, the current must be in Amperes.

Q: What if $V_0$ was not zero?

We would simply use the exact value of $V_0$ in our equation. For example, if $V_0 = 0.5$ V, the equation would be $0.5 = 1.5 - 1 \cdot r$, giving $r = 1.0\ \Omega$.

Q: Can the terminal voltage ever be greater than the EMF?

Yes! But only when the battery is being charged by an external source. In that case, the current direction reverses, and the equation becomes $V = E + Ir$.

Question

To verify Ohm's law, a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current and the following graph is obtained
If $V_0$ is almost zero, then identify the correct statement.

Options:

(A) The emf of the battery is $1.5	ext{ V}$ and its internal resistance is $1.5\ \Omega$

(B) The value of the resistance $R$ is $1.5\ \Omega$

(C) The potential difference across the battery is $1.5	ext{ V}$ when it sends a current of $1000	ext{ mA}$

(D) The emf of the battery is $1.5	ext{ V}$ and the value of $R$ is $1.5\ \Omega$

Sigma Percentile · Accepted Answer

The problem presents us with a classic experimental setup designed to verify Ohm's law and understand the internal characteristics of a battery. We are given a circuit diagram and a corresponding $V-I$ (Voltage-Current) graph. Our goal is to determine the electromotive force (EMF) and the internal resistance of the battery.

### Analyzing the Setup

When a voltmeter is connected in parallel across a battery, it measures the **terminal voltage** ($V$) of the battery. It's crucial to understand that the terminal voltage is not always equal to the battery's EMF ($E$). When a current ($I$) flows through the circuit, the battery's internal resistance ($r$) causes a small voltage drop inside the battery itself.

The relationship between these quantities for a discharging battery is given by the master equation:
$$V = E - Ir$$

This equation tells us that the terminal voltage decreases linearly as the current drawn from the battery increases. This perfectly matches the straight-line graph provided in the problem.

### Decoding the Graph

The given graph plots the terminal voltage $V$ on the y-axis against the current $I$ on the x-axis. We can extract two vital pieces of information from the intercepts of this line.

**1. The y-intercept:**
The graph intersects the y-axis at $V = 1.5	ext{ V}$. At this point, the current $I = 0$. 
Substituting these values into our master equation:
$$1.5 = E - (0)r$$
$$E = 1.5	ext{ V}$$
This gives us the EMF of the battery directly! When no current is drawn, the terminal voltage equals the EMF.

**2. The x-intercept:**
The problem states that $V_0$ is almost zero. Looking at the graph, this near-zero voltage occurs when the current is $1000	ext{ mA}$. 
Before plugging this into our equation, we must ensure our units are consistent. Let's convert milliamperes to amperes:
$$I = 1000	ext{ mA} = 1	ext{ A}$$
Now, substituting $V = 0$, $E = 1.5	ext{ V}$, and $I = 1	ext{ A}$ into the master equation:
$$0 = 1.5 - (1)r$$
$$r = 1.5\ \Omega$$

### Final Conclusion

By analyzing the intercepts of the $V-I$ graph, we have successfully determined both the internal characteristics of the battery. The EMF of the battery is **$1.5	ext{ V}$**, and its internal resistance is **$1.5\ \Omega$**.

Comparing this with the given options, we find that option (a) is the correct statement.

Select Answer:

Visualized Solution

$V = E - I r$

$y$ -intercept

Finding EMF

$x$ -intercept

Finding Internal Resistance

Conclusion

The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup

Decoding the Graph

Final Conclusion

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