The series combination of two batteries, both of the same emf 10 V, but different internal resistanc

Q: Why is the potential difference across the battery zero?

The potential difference across a discharging battery is $V = E - ir$. If the internal voltage drop ($ir$) exactly equals the EMF ($E$), the net potential difference across its terminals becomes zero.

Q: Why do we add the EMFs and internal resistances of the two batteries?

Because the batteries are connected in series, their EMFs add up to give the total driving voltage, and their internal resistances add up to give the total internal resistance of the source.

Q: Can the terminal voltage of a battery be negative?

Yes, if the battery is being charged by a stronger external source, the current direction reverses, and the terminal voltage becomes $V = E + ir$, which is greater than $E$. If it's discharging and $ir > E$, it would be negative, but that usually means another source is dominating the circuit.

Q: Why is the voltage across both parallel resistors the same?

In a parallel circuit, all components are connected across the same two nodes. Therefore, the potential difference between those two nodes is the same for every component connected in parallel.

Q: What if the batteries were connected in parallel?

If the batteries were in parallel, we would have to use the formula for the equivalent EMF of parallel cells: $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$, and the equivalent internal resistance would be $1/r_{eq} = 1/r_1 + 1/r_2$.

Question

The series combination of two batteries, both of the same emf $10	ext{ V}$, but different internal resistance of $20\,\Omega$ and $5\,\Omega$, is connected to the parallel combination of two resistors $30\,\Omega$ and $R\,\Omega$. The voltage difference across the battery of internal resistance $20\,\Omega$ is zero, the value of $R$ (in $\Omega$) is ............... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are looking at a circuit with two main branches. In the bottom branch, two batteries are working together, connected in series. Both have an EMF of $10	ext{ V}$, but they have different internal resistances: $20\,\Omega$ and $5\,\Omega$. In the top branch, we have two resistors connected in parallel: one is $30\,\Omega$ and the other is an unknown $R$.

The total current $i$ flows out of the batteries, splits at the junction into the two resistors, and then recombines. The most crucial piece of information given to us is that the potential difference across the first battery (the one with $20\,\Omega$ internal resistance) is exactly zero.

### The Master Equation
When a battery is discharging, the potential difference $V$ across its terminals is not just its EMF. It drops due to its internal resistance. The formula is:
$$V = E - ir$$

For our first battery, we are told $V_1 = 0$. Let's plug in the values we know:
$$0 = 10 - i(20)$$

Solving this simple equation gives us the total current flowing in the main circuit:
$$20i = 10 \implies i = 0.5	ext{ A}$$

### Finding the Voltage Across the Resistors
Now that we know the total current, we can find the voltage across the parallel combination of resistors. This voltage, let's call it $V_{AB}$, must be equal to the net terminal voltage of the two batteries combined.

The total EMF of the series batteries is $E_1 + E_2 = 10 + 10 = 20	ext{ V}$.
The total internal resistance is $r_1 + r_2 = 20 + 5 = 25\,\Omega$.

So, the net terminal voltage is:
$$V_{AB} = (E_1 + E_2) - i(r_1 + r_2)$$
$$V_{AB} = 20 - 0.5(25)$$
$$V_{AB} = 20 - 12.5 = 7.5	ext{ V}$$

### Final Calculation
This $7.5	ext{ V}$ is the potential difference across both the $30\,\Omega$ resistor and the unknown resistor $R$.

According to Kirchhoff's Current Law, the total current $i$ splits into the two parallel branches:
$$i = i_1 + i_2$$

Using Ohm's law ($i = \frac{V}{R}$) for each branch, we can write:
$$i = \frac{V_{AB}}{R} + \frac{V_{AB}}{30}$$

Let's substitute the values we've found:
$$0.5 = \frac{7.5}{R} + \frac{7.5}{30}$$

We know that $\frac{7.5}{30} = 0.25$. Substituting this back:
$$0.5 = \frac{7.5}{R} + 0.25$$
$$\frac{7.5}{R} = 0.25$$

Finally, solving for $R$:
$$R = \frac{7.5}{0.25} = 30\,\Omega$$

And there we have it! The unknown resistance $R$ is $30\,\Omega$.

Enter Numerical Value:

Visualized Solution

$C i r c u i t S e t u p$

$C o n d i t i o n f or Z er o P o t e n t ia l D i f f er e n ce$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g T o t a l C u r r e n t$

$P o t e n t ia l D r o p A cr oss P a r a l l e l C o mbina t i o n$

$C a l c u l a t in g V_{A B}$

$V a l u e o f V_{A B}$

$C u r r e n t D i s t r ib u t i o n$

$S u b s t i t u t in g f or R$

$S o l v in g f or R$

$F ina l V a l u e o f R$

$W ha t i f ?$

The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup Imagine you are looking at a circuit with two main branches

The Master Equation When a battery is discharging, the potential difference $V$ across its terminals is not just its EMF

Finding the Voltage Across the Resistors Now that we know the total current, we can find the voltage across the parallel combination of resistors

Final Calculation

Similar Questions

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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 Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Zero Potential Difference

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

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

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

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

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

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

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

.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 Value of R

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