A cell of emf 1.1 V and internal resistance 0.5 Ω is connected to a wire of resistance 0.5 Ω. Another cell of the same emf is now connected in series with the intention of increasing the current, but the current in the wire remains the same. The internal resistance of the second cell is:
1 Ω
Understanding the Problem: Cell EMF and Internal Resistance
This problem involves analyzing a simple electrical circuit containing cells, their electromotive force (EMF), and internal resistance. We are given a scenario where a circuit with one cell is modified by adding a second cell in series. The key information is that the current flowing through the external wire remains unchanged despite adding the second cell.
We need to use Ohm's Law for the entire circuit, which relates the total EMF, total resistance (external plus internal), and the current.
Case 1: The Initial Circuit with One Cell
Initially, a single cell is connected to a wire (external resistance). Let's denote the parameters of this cell and the wire:
- EMF of the first cell, $E_1 = 1.1 \text{ V}$
- Internal resistance of the first cell, $r_1 = 0.5 \text{ Ω}$
- Resistance of the wire (external resistance), $R = 0.5 \text{ Ω}$
The current $I_1$ flowing in the circuit can be calculated using the formula for current in a circuit with internal resistance:
\$I_1 = \frac{\text{Total EMF}}{\text{External Resistance} + \text{Total Internal Resistance}}\$
In this case, the total EMF is $E_1$ and the total internal resistance is $r_1$.
\$I_1 = \frac{E_1}{R + r_1}\$
Substituting the given values:
\$I_1 = \frac{1.1 \text{ V}}{0.5 \text{ Ω} + 0.5 \text{ Ω}}\$
\$I_1 = \frac{1.1 \text{ V}}{1.0 \text{ Ω}}\$
\$I_1 = 1.1 \text{ A}\$
So, the initial current in the wire is 1.1 A.
Case 2: Adding a Second Cell in Series
A second cell with the same EMF ($E_2 = 1.1 \text{ V}$) is connected in series with the first cell. Let the internal resistance of this second cell be $r_2$. The same wire (external resistance $R = 0.5 \text{ Ω}$) is still in the circuit.
When cells are connected in series such that their polarities add up (positive terminal of one connected to the negative terminal of the next), the total EMF is the sum of their individual EMFs, and the total internal resistance is the sum of their individual internal resistances.
- Total EMF, $E_{total} = E_1 + E_2$
- Total internal resistance, $r_{total} = r_1 + r_2$
Substituting the values:
- $E_{total} = 1.1 \text{ V} + 1.1 \text{ V} = 2.2 \text{ V}$
- $r_{total} = 0.5 \text{ Ω} + r_2$
The current $I_2$ in this new circuit is given by:
\$I_2 = \frac{E_{total}}{R + r_{total}}\$
\$I_2 = \frac{2.2 \text{ V}}{0.5 \text{ Ω} + (0.5 \text{ Ω} + r_2)}\$
\$I_2 = \frac{2.2 \text{ V}}{1.0 \text{ Ω} + r_2}\$
The Condition: Current Remains the Same
The problem states that the current in the wire remains the same after adding the second cell. This means $I_2 = I_1$.
We calculated $I_1 = 1.1 \text{ A}$. So, $I_2$ must also be 1.1 A.
Set the expression for $I_2$ equal to 1.1 A:
\$\frac{2.2}{1.0 + r_2} = 1.1\$
Solving for the Internal Resistance of the Second Cell ($r_2$)
Now, we need to solve this equation for $r_2$.
Multiply both sides by &$(1.0 + r_2)$:
\$2.2 = 1.1 \times (1.0 + r_2)\$
Distribute 1.1 on the right side:
\$2.2 = 1.1 \times 1.0 + 1.1 \times r_2\$
\$2.2 = 1.1 + 1.1 r_2\$
Subtract 1.1 from both sides:
\$2.2 - 1.1 = 1.1 r_2\$
\$1.1 = 1.1 r_2\$
Divide both sides by 1.1:
\$r_2 = \frac{1.1}{1.1}\$
\$r_2 = 1.0 \text{ Ω}\$
The internal resistance of the second cell is 1.0 Ω.
Conclusion
Based on the calculation, the internal resistance of the second cell must be 1 Ω for the current in the wire to remain the same when it is connected in series with the first cell of the same EMF.
This happens because while the total EMF is doubled (from 1.1 V to 2.2 V), the total resistance of the circuit (external + total internal) must also double for the current to remain constant (Current $= \text{EMF}/\text{Resistance}$). If the external resistance is constant, the total internal resistance must increase appropriately.
| Circuit Configuration | Total EMF | Total Internal Resistance | External Resistance | Total Resistance | Current |
|---|---|---|---|---|---|
| One Cell | $1.1 \text{ V}$ | $0.5 \text{ Ω}$ | $0.5 \text{ Ω}$ | $0.5 + 0.5 = 1.0 \text{ Ω}$ | $1.1 / 1.0 = 1.1 \text{ A}$ |
| Two Cells (Series) | $1.1 + 1.1 = 2.2 \text{ V}$ | $0.5 + r_2$ | $0.5 \text{ Ω}$ | $0.5 + (0.5 + r_2) = 1.0 + r_2$ | $2.2 / (1.0 + r_2)$ |
For current to be same:
\$1.1 = \frac{2.2}{1.0 + r_2}\$
\$1.0 + r_2 = \frac{2.2}{1.1}\$
\$1.0 + r_2 = 2.0\$
\$r_2 = 2.0 - 1.0\$
\$r_2 = 1.0 \text{ Ω}\$
Revision Table: Cell EMF and Resistance
| Concept | Description | Formula (for a simple circuit) |
|---|---|---|
| EMF ($E$) | The potential difference across the terminals of a cell when no current is flowing (open circuit). It is the energy provided by the cell per unit charge. | N/A |
| Internal Resistance ($r$) | Resistance offered by the electrolyte and electrodes of the cell itself to the flow of current. | N/A |
| External Resistance ($R$) | Total resistance of the components in the circuit outside the cell. | N/A |
| Ohm's Law (for complete circuit) | Relates total EMF, total resistance, and current. | $I = \frac{E}{R+r}$ (for single cell) $I = \frac{E_{total}}{R_{external} + R_{internal\_total}}$ (for multiple cells) |
| Cells in Series | Connecting cells positive to negative. Total EMF adds up. Total internal resistance adds up. | $E_{total} = E_1 + E_2 + ...$ $r_{total} = r_1 + r_2 + ...$ |
| Potential Difference across terminals ($V$) | The voltage available across the external circuit terminals when current is flowing. | $V = E - Ir$ $V = IR$ |
Additional Information: Cell Combinations and Current
Connecting cells in series is often done to increase the total voltage (EMF) available to the circuit. This usually leads to a higher current, assuming the external resistance is significant compared to the internal resistances.
However, the internal resistance of the cells also adds up in series. In this specific problem, the current remained constant. This indicates a balance between the increase in total EMF and the increase in total internal resistance relative to the external resistance.
If the external resistance ($R$) is much larger than the total internal resistance ($r_{total}$), the current is approximately $I \approx E_{total} / R$. In this case, adding cells in series would indeed increase the current proportionally to the total EMF.
If the external resistance is comparable to or smaller than the internal resistances, the increase in total internal resistance significantly impacts the current. As seen in this problem, doubling the EMF by adding a cell requires doubling the total resistance ($R + r_{total}$) to keep the current constant. Since $R$ is fixed, the total internal resistance $r_{total}$ must increase such that &$(R + r_{total})_{new} = 2 \times (R + r_{total})_{old}&$.
Connecting cells in parallel is another configuration, typically used to increase the total current capacity or reduce the effective internal resistance, especially when powering low-resistance loads or when cells have different capacities.
Important Questions from Current Electricity
The current through a 4/3 Ω external resistance connected to a parallel combination of two cells of 2 V and 1 V emf and internal resistances of 1 Ω and 2 Ω respectively is:
A metallic wire of uniform area of cross-section has a resistance R, resistivity ρ, and power rating P at V volts. The wire is uniformly stretched to reduce the radius to half the original radius. The values of resistance, resistivity, and power rating at V volts are now denoted by R', ρ', and P' respectively. The corresponding values are correctly related as:
P, Q, R, and S are four wires of resistances 3 Ω, 3 Ω, 3 Ω, and 4 Ω, respectively. They are connected to form the four arms of a Wheatstone bridge circuit. The resistance with which S must be shunted in order that the bridge may be balanced is:
The magnetic moment of a thin bar magnet is 'M'. If it is bent into a semicircular form, its new magnetic moment will be:
The current flowing through the two bulbs marked as 60W, 240V each when connected in series with a 240V source is:
