In an electric circuit, a cell of certain EMF provides a potential difference of 1.25 V across a loa

Q: Why does the terminal voltage change when we change the load resistance?

The cell has an internal resistance. When you decrease the load resistance, more current flows through the circuit. This increased current causes a larger voltage drop across the internal resistance ($Ir$), leaving less voltage available at the terminals ($V = \varepsilon - Ir$).

Q: Can we just use $V = IR$ and ignore internal resistance?

No! If the cell were ideal (zero internal resistance), the terminal voltage would always equal the EMF, regardless of the load. The fact that the voltage changes from $1.25\text{ V}$ to $1\text{ V}$ is the absolute proof that internal resistance exists and must be accounted for.

Q: Why did we write $\varepsilon = V(1 + r/R_L)$ instead of just calculating the currents first?

It's a mathematical shortcut! While you can calculate the currents first ($I_1 = 1.25/5$ and $I_2 = 1/2$) and use $\varepsilon = V + Ir$, writing it as a single ratio formula often makes the algebraic substitution cleaner and less prone to decimal errors.

Q: What if the internal resistance was zero?

If $r = 0$, then the terminal voltage $V$ would always be exactly equal to the EMF $\varepsilon$. In this problem, $V$ would be $1.25\text{ V}$ in both cases, which contradicts the given data.

Q: I got $r = 1 \Omega$, but how do I know which equation to plug it back into?

You can plug it into either equation! Since both equations describe the exact same cell, substituting $r = 1$ into the first case ($1.25 \times (1 + 1/5)$) or the second case ($1 \times (1 + 1/2)$) will give you the exact same EMF of $1.5\text{ V}$. Try it yourself to verify!

Question

In an electric circuit, a cell of certain EMF provides a potential difference of $1.25$ V across a load resistance of $5 \Omega$. However, it provides a potential difference of $1$ V across a load resistance of $2 \Omega$. The emf of the cell is given by $\frac{x}{10}$ V. Then, the value of $x$ is ......... .

Sigma Percentile · Accepted Answer

### The Tale of the Hidden Resistance: Unmasking the Cell's True EMF

Imagine you are holding a standard battery. It proudly claims to be a $1.5	ext{ V}$ source. But when you connect it to a heavy load, like a powerful motor, the voltage it actually delivers seems to magically drop. Where did the missing voltage go?

This classic phenomenon is the heart of our problem. A real-world cell is not just a perfect voltage source; it has its own internal struggles, known as **internal resistance** ($r$). When current flows out of the cell, it must fight its way through this internal resistance, causing a voltage drop *inside* the cell itself.

### The Master Equation

Let's formalize this physical intuition. The actual voltage available to the outside world is called the **terminal voltage** ($V$). It is simply the cell's true Electromotive Force (EMF, $\varepsilon$) minus the internal voltage drop ($Ir$).

$$V = \varepsilon - Ir$$

We also know from Ohm's law that the terminal voltage is what drives the current through the external load resistance ($R_L$). Therefore, $V = I R_L$, which means the current is $I = \frac{V}{R_L}$.

If we substitute this expression for current back into our first equation, we can eliminate $I$ entirely:

$$\varepsilon = V + \left(\frac{V}{R_L}ight) r$$

Factoring out the terminal voltage $V$, we arrive at a beautiful, highly useful master equation:

$$\varepsilon = V \left(1 + \frac{r}{R_L}ight)$$

This equation is incredibly powerful because it directly links the three things we can easily measure or change: the terminal voltage, the load resistance, and the internal properties of the cell.

### A Tale of Two Scenarios

The problem provides us with two distinct experimental scenarios using the exact same cell. This is perfect because we have two unknowns: the EMF ($\varepsilon$) and the internal resistance ($r$). Two scenarios mean two equations!

**Scenario 1:** The cell is connected to a $5 \Omega$ load, and the terminal voltage is $1.25	ext{ V}$. Plugging this into our master equation gives:

$$\varepsilon = 1.25 \left(1 + \frac{r}{5}ight)$$

**Scenario 2:** The load is swapped for a smaller $2 \Omega$ resistor, and the terminal voltage drops to $1	ext{ V}$. This makes perfect physical sense! A smaller load draws more current, which increases the internal voltage drop ($Ir$), leaving less voltage for the outside world. Our second equation is:

$$\varepsilon = 1 \left(1 + \frac{r}{2}ight)$$

### The Grand Equating

Since we are using the same cell in both scenarios, its true EMF ($\varepsilon$) remains absolutely constant. Therefore, we can confidently equate the right-hand sides of our two equations:

$$1.25 \left(1 + \frac{r}{5}ight) = 1 \left(1 + \frac{r}{2}ight)$$

Now, it's just a matter of careful algebra. Let's expand the brackets. Remember that $\frac{1.25}{5} = 0.25$:

$$1.25 + 0.25r = 1 + 0.5r$$

Let's group the constant terms on the left and the $r$ terms on the right:

$$1.25 - 1 = 0.5r - 0.25r$$

$$0.25 = 0.25r$$

Dividing both sides by $0.25$, we find the hidden internal resistance:

$$r = 1 \Omega$$

### The Final Reveal

With the internal resistance unmasked, finding the true EMF is a breeze. We can substitute $r = 1 \Omega$ back into either of our original scenario equations. Let's use the second one because the math is slightly simpler:

$$\varepsilon = 1 \left(1 + \frac{1}{2}ight)$$

$$\varepsilon = 1 	imes 1.5 = 1.5	ext{ V}$$

We have successfully found the cell's EMF! But the problem asks for a specific format. It states that the EMF is given by the expression $\frac{x}{10}	ext{ V}$.

Equating our result to this expression:

$$\frac{x}{10} = 1.5$$

Multiplying both sides by $10$, we get our final, triumphant answer:

$$x = 15$$

This problem is a fantastic reminder that in physics, what you measure on the outside (terminal voltage) is often just a shadow of the true potential (EMF) hidden within, masked by the system's own internal constraints.

Enter Numerical Value:

Visualized Solution

Visualizing the Circuit

The Master Equation

Setting up Case 1

Setting up Case 2

Equating the EMFs

Solving for Internal Resistance

Calculating the EMF

Finding the Value of x

Physical Reflection

The Sigma Insight: Cells, EMF, and Internal Resistance

The Tale of the Hidden Resistance

The Master Equation

A Tale of Two Scenarios

The Grand Equating

The Final Reveal

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Enter Numerical Value:

Visualized Solution

Visualizing the Circuit

The Master Equation

Setting up Case 1

Setting up Case 2

Equating the EMFs

Solving for Internal Resistance

Calculating the EMF

Finding the Value of x

Physical Reflection

The Sigma Insight: Cells, EMF, and Internal Resistance

The Tale of the Hidden Resistance

The Master Equation

A Tale of Two Scenarios

The Grand Equating

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