When two identical batteries of internal resistance 1 ohm each are connected in series across a resi

Q: Why does the EMF remain $E$ in the parallel combination?

In a parallel combination of identical batteries, the potential difference across the branches is the same. Since they are identical, they don't drive current into each other, and the net potential difference available to the external circuit is just the EMF of a single battery.

Q: Why do we add internal resistances in series but use the parallel formula for parallel?

Internal resistance behaves exactly like a regular resistor connected in series with the ideal battery. When batteries are in series, their internal resistors are in series. When in parallel, the current splits between the branches, so the internal resistors are in parallel.

Q: Can I expand the squares to solve the quadratic equation?

Yes, you can expand $(R+2)^2$ and $(R+0.5)^2$ to form a quadratic equation in $R$. However, since both sides are perfect squares, taking the square root is a much faster and less error-prone mathematical shortcut.

Q: What if the batteries were not identical?

If the batteries had different EMFs or internal resistances, you would have to use Kirchhoff's laws or equivalent battery formulas (like Millman's theorem) to find the net EMF and net internal resistance.

Q: Does the value of $E$ matter for the final answer?

No, the EMF $E$ cancels out when we take the ratio of the powers. This means the required external resistance $R$ depends only on the internal resistances and the given power ratio, not on the actual voltage of the batteries.

Question

When two identical batteries of internal resistance $1\Omega$ each are connected in series across a resistor $R$, the rate of heat produced in $R$ is $J_1$. When the same batteries are connected in parallel across $R$, the rate is $J_2$. If $J_1 = 2.25 J_2$ then the value of $R$ in $\Omega$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an electrical engineer tasked with extracting power from a set of batteries. You have two identical batteries, each with an electromotive force (EMF) of $E$ and an internal resistance of $1\,\Omega$.

You decide to test two different configurations: connecting them in series and connecting them in parallel. In both cases, you hook them up to the same external resistor $R$.

Our goal is to find the exact value of $R$ given that the rate of heat produced in the series circuit is $2.25$ times the rate of heat produced in the parallel circuit.

### The Master Equation

The rate of heat produced in a resistor is simply the electrical power dissipated by it. According to Joule's law of heating, this power is given by:

$$P = i^2 R$$

To find the current $i$ flowing through the external resistor, we use Ohm's law for a complete circuit. The current is the total equivalent EMF divided by the total equivalent resistance (external plus internal):

$$i = \frac{E_{eq}}{R + r_{eq}}$$

### Evaluating the Series Circuit

Let's look at the series combination first. When batteries are connected in series, their EMFs add up, and their internal resistances add up as well.

$$E_{eq} = E + E = 2E$$

$$r_{eq} = 1\,\Omega + 1\,\Omega = 2\,\Omega$$

The current $i_1$ in the series circuit is:

$$i_1 = \frac{2E}{R + 2}$$

Therefore, the rate of heat produced, $J_1$, is:

$$J_1 = \left(\frac{2E}{R + 2}ight)^2 R$$

### Evaluating the Parallel Circuit

Now, let's analyze the parallel combination. When identical batteries are connected in parallel, the equivalent EMF remains the same as a single battery. However, their internal resistances are now in parallel, which halves the total internal resistance.

$$E_{eq} = E$$

$$r_{eq} = \frac{1 	imes 1}{1 + 1} = 0.5\,\Omega$$

The current $i_2$ in the parallel circuit is:

$$i_2 = \frac{E}{R + 0.5}$$

The rate of heat produced, $J_2$, is:

$$J_2 = \left(\frac{E}{R + 0.5}ight)^2 R$$

### Final Calculation

We are given a crucial piece of information: $J_1 = 2.25 J_2$. Let's substitute our expressions into this equation.

$$\left(\frac{2E}{R + 2}ight)^2 R = 2.25 \left(\frac{E}{R + 0.5}ight)^2 R$$

Notice how elegantly the $E^2$ and $R$ terms cancel out from both sides. We are left with a purely algebraic equation in terms of $R$:

$$\frac{4}{(R + 2)^2} = \frac{2.25}{(R + 0.5)^2}$$

**Here is a pro-tip:** Do not expand the squares! That will lead to a messy quadratic equation. Instead, simply take the square root of both sides.

$$\frac{2}{R + 2} = \frac{1.5}{R + 0.5}$$

Now, cross-multiply to solve for $R$:

$$2(R + 0.5) = 1.5(R + 2)$$

$$2R + 1 = 1.5R + 3$$

$$0.5R = 2$$

$$R = 4\,\Omega$$

The external resistance required to satisfy the given condition is exactly **$4\,\Omega$**.

Enter Numerical Value:

Visualized Solution

$Circuit Configurations$

$Power and Current Formulas$

$Series Combination$

$Parallel Combination$

$Equating the Heat Rates$

$Solving for R$

$Final Calculation$

$Conclusion$

The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup

The Master Equation

Evaluating the Series Circuit

Evaluating the Parallel Circuit

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Configurations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Power and Current Formulas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Series Combination

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Combination

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating the Heat Rates

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion