Internal resistance of cells and Kirchhoff’s laws
Internal Resistance of Cells
When an electric current flows through a cell, it encounters resistance not only from the external circuit but also from the electrolyte and electrodes within the cell itself. This inherent opposition to current flow within the cell is called internal resistance, denoted by 'r'.
The magnitude of internal resistance depends on several factors:
- Nature of the electrolyte: The conductivity of the electrolyte is a primary factor. Highly conductive electrolytes offer less internal resistance.
- Distance between electrodes: A greater distance between the electrodes increases the path length for the ions, thus increasing resistance.
- Area of electrodes: A larger surface area of the electrodes in contact with the electrolyte reduces the resistance.
- Concentration of the electrolyte: For electrolytes, higher concentration generally leads to higher conductivity and thus lower internal resistance.
- Temperature: Temperature affects the conductivity of the electrolyte. For most electrolytes, conductivity increases with temperature, leading to a decrease in internal resistance.
EMF and Terminal Voltage
An ideal cell, when delivering no current, has its positive and negative terminals at a potential difference equal to its electromotive force (EMF), denoted by 'ε'. The EMF is the maximum potential difference a cell can provide and is a characteristic property of the cell itself, independent of the external circuit.
However, when a cell is connected to an external circuit and delivers a current 'I', there is a voltage drop across its internal resistance. This voltage drop is given by $V_{drop} = I \times r$.
The potential difference available across the terminals of the cell in the external circuit is called the terminal voltage (V). It is calculated as the EMF minus the voltage drop due to internal resistance:
$$V = \epsilon - Ir$$
Case 1: Cell is supplying current to an external resistance (Discharging)
In this case, the terminal voltage is less than the EMF: $V = \epsilon - Ir$. The current flowing through the circuit is $I = \frac{\epsilon}{R + r}$, where R is the external resistance.
Case 2: Cell is being charged by an external source
When a cell is being charged, an external voltage source forces current into the cell, opposing its natural tendency to discharge. In this scenario, the terminal voltage of the cell is greater than its EMF. The external voltage must overcome both the EMF and the internal resistance drop. The terminal voltage is then:
$$V = \epsilon + Ir$$
The current flowing into the cell during charging is $I = \frac{V_{external} - \epsilon}{r}$, where $V_{external}$ is the voltage of the external charging source.
Case 3: Cell is on open circuit
When the terminals of the cell are not connected to any external circuit (i.e., the circuit is open), no current flows ($I = 0$). In this situation, the terminal voltage is equal to the EMF:
$$V = \epsilon - (0)r = \epsilon$$
Determining Internal Resistance
Internal resistance can be determined experimentally. Consider a cell with EMF ε and internal resistance r, connected to an external resistance R. The current in the circuit is $I = \frac{\epsilon}{R + r}$. The terminal voltage is $V = IR = \frac{\epsilon R}{R + r}$.
We can also write $V = \epsilon - Ir$. Rearranging this, we get $Ir = \epsilon - V$. Therefore, $r = \frac{\epsilon - V}{I}$.
Substituting $I = \frac{V}{R}$ (since V is the voltage across the external resistance R), we get:
$$r = \frac{\epsilon - V}{V/R} = R \left(\frac{\epsilon}{V} - 1\right)$$
This formula allows us to calculate the internal resistance if we can measure the terminal voltage (V) when a known external resistance (R) is connected and the cell is supplying current.
A common experimental method involves measuring the terminal voltage with two different external resistances, $R_1$ and $R_2$.
When external resistance is $R_1$, terminal voltage is $V_1$. Current $I_1 = \frac{V_1}{R_1}$.
$$V_1 = \epsilon - I_1 r \implies V_1 = \epsilon - \frac{V_1}{R_1} r$$
When external resistance is $R_2$, terminal voltage is $V_2$. Current $I_2 = \frac{V_2}{R_2}$.
$$V_2 = \epsilon - I_2 r \implies V_2 = \epsilon - \frac{V_2}{R_2} r$$
From these two equations, we can eliminate ε and solve for r. Or, more practically, if we know the EMF (perhaps by measuring terminal voltage when R is very large, approaching infinity, or from the cell's specifications), we can use the formula $r = R \left(\frac{\epsilon}{V} - 1\right)$.
Combination of Cells:
When cells are connected in series, their EMFs add up and their internal resistances add up:
Equivalent EMF, $\epsilon_{eq} = \epsilon_1 + \epsilon_2 + ...$
Equivalent internal resistance, $r_{eq} = r_1 + r_2 + ...$
When cells are connected in parallel, the equivalent EMF and internal resistance calculation is more complex and depends on whether the cells have identical EMFs and internal resistances. For identical cells (EMF ε, resistance r) connected in parallel, the equivalent EMF is ε and the equivalent internal resistance is $r_{eq} = \frac{r}{n}$, where n is the number of cells.
Kirchhoff’s Laws
Kirchhoff's laws are fundamental principles used to analyze complex electrical circuits containing multiple loops and junctions. They are extensions of Ohm's law and are based on the conservation of charge and energy.
1. Kirchhoff’s Junction Rule (First Law / Current Law)
This law is based on the principle of conservation of electric charge. It states that for any junction (or node) in an electrical circuit, the sum of currents entering the junction must equal the sum of currents leaving the junction.
Mathematically:
$$\sum I_{in} = \sum I_{out}$$
Or, considering currents entering as positive and currents leaving as negative:
$$\sum I = 0 \quad \text{(at a junction)}$$
Example:
Consider a junction where three wires meet. If currents $I_1$ and $I_2$ are flowing into the junction, and current $I_3$ is flowing out, then according to Kirchhoff's junction rule:
$$I_1 + I_2 = I_3$$
Explanation: A junction is a point where three or more conductors meet. Charge cannot accumulate at a junction. Therefore, the rate at which charge enters the junction must be equal to the rate at which charge leaves it. This is analogous to water flow in pipes; the total water flowing into a junction must equal the total water flowing out.
2. Kirchhoff’s Loop Rule (Second Law / Voltage Law)
This law is based on the principle of conservation of energy. It states that for any closed loop in an electrical circuit, the algebraic sum of the potential differences (voltages) around the loop must be zero.
Mathematically:
$$\sum V = 0 \quad \text{(around a closed loop)}$$
Explanation: If you start at any point in a closed circuit and traverse the loop, returning to the starting point, the net change in electric potential must be zero. This means that any energy supplied by sources (like cells) within the loop is entirely dissipated by resistances or other components in the loop.
When applying the loop rule, we need to consider the sign conventions for voltage changes:
- Resistors: If traversing a resistor in the direction of the current, there is a voltage drop ($-\text{IR}$). If traversing against the current, there is a voltage rise ($+\text{IR}$).
- Cells (EMF sources): If traversing a cell from the negative terminal to the positive terminal, there is a voltage rise ($+\epsilon$). If traversing from the positive terminal to the negative terminal, there is a voltage drop ($-\epsilon$).
Example:
Consider a simple circuit with a single loop containing two cells ($\epsilon_1, r_1$ and $\epsilon_2, r_2$) and an external resistance R. Let's assume a current I flows clockwise.
Starting from the bottom-left corner and moving clockwise:
1. Across $r_1$: Moving with current I, voltage drop is $-Ir_1$.
2. Across $\epsilon_1$: Moving from negative to positive, voltage rise is $+\epsilon_1$.
3. Across R: Moving with current I, voltage drop is $-IR$.
4. Across $\epsilon_2$: Moving from positive to negative, voltage drop is $-\epsilon_2$.
5. Across $r_2$: Moving with current I, voltage drop is $-Ir_2$.
Applying the loop rule:
$$-\text{Ir}_1 + \epsilon_1 - \text{IR} - \epsilon_2 - \text{Ir}_2 = 0$$
Rearranging to solve for I:
$$\epsilon_1 - \epsilon_2 = I(R + r_1 + r_2)$$
$$I = \frac{\epsilon_1 - \epsilon_2}{R + r_1 + r_2}$$
Note: The sign of the current indicates its direction. If the calculated I is positive, the assumed clockwise direction is correct. If negative, the current flows counter-clockwise.
- Identify Junctions and Loops: Mark all junctions and at least one independent loop for each part of the circuit.
- Assign Current Directions: Assume a direction for the current in each branch. It's okay if your assumption is wrong; the calculated current will come out negative, indicating the actual direction is opposite.
- Apply Junction Rule: Write down the equation for each junction (n-1 junctions are independent for a circuit with n junctions).
- Apply Loop Rule: For each independent loop, write down the voltage equation, carefully applying sign conventions.
- Solve the System of Equations: You will have a system of linear equations. Solve them simultaneously to find the unknown currents.
Applications of Kirchhoff's Laws
Kirchhoff's laws are essential for analyzing circuits that cannot be simplified using series and parallel combinations alone. This includes:
- Wheatstone bridge circuits.
- Potentiometer and meter bridge circuits.
- Complex networks with multiple power sources.
- Analyzing the behavior of cells with internal resistance in various configurations.
Example Problem:
Consider a circuit with two loops. Loop 1 contains a 2V cell with internal resistance 1Ω and a 3Ω resistor. Loop 2 contains a 3V cell with internal resistance 1Ω and the same 3Ω resistor, but connected such that the cells oppose each other. Find the current in each part of the circuit.
Let's draw this. Assume two loops. A common junction connects the 3Ω resistor to a point above. To the left of the 3Ω resistor is the 2V cell (positive terminal up). To the right of the 3Ω resistor is the 3V cell (positive terminal down, opposing the 2V cell).
Let $I_1$ be the current flowing from the 2V cell through the 3Ω resistor, and $I_2$ be the current flowing from the 3V cell through the 3Ω resistor. The current in the 3Ω resistor will be the difference between these two.
Let's define currents more precisely. Let $I_1$ flow clockwise in the left loop (through 2V, 1Ω, 3Ω). Let $I_2$ flow clockwise in the right loop (through 3V, 1Ω, 3Ω). The junction rule at the top connection point of the 3Ω resistor states that the current leaving is $I_1 + I_2$ (if both $I_1$ and $I_2$ are assumed to flow towards the junction from their respective loops) or something similar depending on how you define loops and currents.
A better approach: Let $I$ be the current flowing from the 2V cell upwards through the 3Ω resistor. Let $I'$ be the current flowing from the 3V cell downwards through the 3Ω resistor. The net current through the 3Ω resistor is $I - I'$.
Loop 1 (Left loop): Starting from the bottom of the 2V cell, going up:
$$+2V - (1\Omega)I - (3\Omega)I = 0$$
$$2 = 4I \implies I = 0.5 A$$
The current $I$ flows upwards through the 3Ω resistor.
Loop 2 (Right loop): Starting from the bottom of the 3V cell, going up:
$$+3V - (1\Omega)I' - (3\Omega)I' = 0$$
$$3 = 4I' \implies I' = 0.75 A$$
The current $I'$ flows downwards through the 3Ω resistor.
Now, let's reconsider the setup. The problem states the cells oppose each other *through* the 3Ω resistor. This implies the 3Ω resistor is common to both loops.
Let's assume the 3Ω resistor is in the middle. To its left is the 2V cell and its 1Ω internal resistance. To its right is the 3V cell and its 1Ω internal resistance. Let the current through the 3Ω resistor be $I_{net}$.
Consider the left loop: Let current $I_L$ flow clockwise. The 3Ω resistor is part of this loop. Let's assume $I_L$ flows from left to right through the 3Ω resistor.
Starting from the bottom of the 2V cell:
$$+2V - (1\Omega)I_L - (3\Omega)I_L = 0$$
$$2 = 4I_L \implies I_L = 0.5 A$$
So, current of 0.5 A flows from left to right through the 3Ω resistor.
Consider the right loop: Let current $I_R$ flow clockwise. The 3Ω resistor is part of this loop. Let's assume $I_R$ flows from right to left through the 3Ω resistor.
Starting from the bottom of the 3V cell:
$$+3V - (1\Omega)I_R - (3\Omega)I_R = 0$$
$$3 = 4I_R \implies I_R = 0.75 A$$
So, current of 0.75 A flows from right to left through the 3Ω resistor.
The net current through the 3Ω resistor is the difference between the current flowing from left to right ($I_L = 0.5$ A) and the current flowing from right to left ($I_R = 0.75$ A). Since $I_R$ is larger and flows in the opposite direction, the net current will be from right to left.
$$I_{net} = I_R - I_L = 0.75 \text{ A} - 0.5 \text{ A} = 0.25 \text{ A}$$
The net current of 0.25 A flows from right to left through the 3Ω resistor.
This example highlights the importance of carefully defining loops, current directions, and applying the sign conventions consistently.