Five identical cells each of internal resistance 1 ohm and emf 5V are connected in series and in par

Q: Why is the equivalent EMF in parallel just $E$ and not $5E$?

In a parallel combination of identical cells, the potential difference across the terminals is the same as that of a single cell because they are all connected across the same two nodes. They share the current load, but the voltage remains $E$.

Q: Why do we add the external resistance $R$ to the equivalent internal resistance?

The external resistance $R$ is connected in series with the entire battery combination (whether the cells themselves are in series or parallel). Therefore, their resistances add up.

Q: What if the cells were not identical?

If the cells had different EMFs or internal resistances, you would have to use Kirchhoff's laws or the general parallel combination formula: $E_{eq} = \frac{\sum (E_i/r_i)}{\sum (1/r_i)}$.

Q: Is the shortcut $R=r$ always true for any number of cells?

Yes! For $n$ identical cells, equating the series and parallel currents always simplifies to $R = r$. It's a fantastic result to memorize.

Q: Why did we cancel the 5 in the numerator?

It's just a mathematical simplification to make the cross-multiplication easier. You could cross-multiply directly, and you would still get the exact same answer!

Question

Five identical cells each of internal resistance $1\Omega$ and emf $5	ext{V}$ are connected in series and in parallel with an external resistance $R$. For what value of $R$, current in series and parallel combination will remain the same ?

Options:

(A) $1\Omega$

(B) $25\Omega$

(C) $5\Omega$

(D) $10\Omega$

Sigma Percentile · Accepted Answer

### The Beauty of Symmetry in Circuits

Imagine you have a handful of identical batteries. You can connect them in a long chain (series) to boost the voltage, or you can connect them side-by-side (parallel) to share the load. In this fascinating problem, we are asked to find a magical value for an external resistance $R$ such that the current flowing through it is exactly the same, regardless of whether our five batteries are in series or parallel. Let's break down the physics behind this.

### Analyzing the Parallel Combination

First, let's look at the parallel configuration. We have $n = 5$ identical cells, each with an electromotive force (EMF) $e = 5	ext{ V}$ and an internal resistance $r = 1\,\Omega$.

When identical cells are connected in parallel, the equivalent EMF of the entire combination is simply equal to the EMF of a single cell. They don't boost the voltage; they just share the effort.

$$V_p = e = 5	ext{ V}$$

However, their internal resistances are also in parallel. The equivalent internal resistance of $5$ identical resistors $r$ in parallel is $r/5$. This entire battery pack is then connected in series with our external resistor $R$. Therefore, the total resistance of the parallel circuit is:

$$R_p = R + \frac{r}{5} = R + \frac{1}{5}\,\Omega$$

### Analyzing the Series Combination

Now, let's rewire those same five cells into a series chain. In a series circuit, the voltages stack up. The net potential is the sum of all the individual EMFs.

$$V_s = e + e + e + e + e = 5e = 25	ext{ V}$$

Similarly, the internal resistances also stack up. The equivalent internal resistance is $5r$. Adding our external resistance $R$, the total resistance of the series circuit becomes:

$$R_s = R + 5r = R + 5\,\Omega$$

### The Master Equation

The core condition of the problem is that the current in both configurations must be identical. According to Ohm's Law, the current $I$ is the total voltage divided by the total resistance ($I = V/R$). Let's equate the currents for both cases:

$$I_p = I_s$$

$$\frac{V_p}{R_p} = \frac{V_s}{R_s}$$

### Final Calculation

Now, we simply substitute the expressions we derived into our master equation:

$$\frac{5}{R + \frac{1}{5}} = \frac{25}{R + 5}$$

To make the algebra cleaner, we can divide both numerators by $5$:

$$\frac{1}{R + 0.2} = \frac{5}{R + 5}$$

Cross-multiplying gives us a straightforward linear equation:

$$R + 5 = 5(R + 0.2)$$

$$R + 5 = 5R + 1$$

Rearranging the terms to isolate $R$:

$$4R = 4$$

$$R = 1\,\Omega$$

### The Grand Takeaway

We found that $R = 1\,\Omega$, which happens to be exactly equal to the internal resistance $r$ of a single cell. This is not a coincidence!

**For any number $n$ of identical cells, the current in a series combination will equal the current in a parallel combination if and only if the external resistance $R$ is exactly equal to the internal resistance $r$ of a single cell.**

This is a brilliant shortcut to keep in your arsenal for competitive exams. If you spot this setup, you can instantly write down $R = r$ without doing any algebra!

Animated Solution for Physics - Current Electricity: Five identical cells each of internal resistance $1Ω$ and emf $5 V$ are connected in series and in parallel with an external resistance $R$ . For what value of $R$ , current in series and parallel combination will remain the same ?

Select Answer:

Visualized Solution

Circuit Configurations

Parallel Combination: Equivalent EMF

Parallel Combination: Equivalent Resistance

Series Combination: Equivalent EMF

Series Combination: Equivalent Resistance

Equating the Currents

Substituting Values

Solving for $R$

The General Case

The Sigma Insight: Cells, EMF, and Internal Resistance

The Beauty of Symmetry in Circuits

Analyzing the Parallel Combination

Analyzing the Series Combination

The Master Equation

Final Calculation

The Grand Takeaway

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