Current Electricity class 12 notes
A Comprehensive Guide to Unit II: Current Electricity (07 marks)
Introduction to Current Electricity
Current electricity deals with the flow of electric charges through conductors. It forms the basis of almost all electrical and electronic devices we use today. In this unit, we will explore the fundamental concepts, laws, and applications of current electricity.
Current electricity is distinct from static electricity in that it involves charges in motion rather than at rest. The continuous flow of these charges enables the transfer of energy across distances, powering everything from simple light bulbs to complex computer systems.
Electric Current
Electric current is defined as the rate of flow of electric charges through a conductor. Mathematically, it is expressed as:
Where:
- \(I\) is the electric current (in amperes, A)
- \(dQ\) is the amount of electric charge (in coulombs, C)
- \(dt\) is the time interval (in seconds, s)
The SI unit of electric current is the ampere (A), which is defined as the flow of one coulomb of charge per second:
Current can be of two types:
- Direct Current (DC): The flow of electric charges is unidirectional. Examples include batteries and solar cells.
- Alternating Current (AC): The direction of flow of electric charges reverses periodically. This is the type of electricity supplied to our homes.
Numerical Example:
A constant current of 2A flows through a conductor. Calculate the amount of charge flowing through the conductor in 5 minutes.
Solution:
Given:
- Current, \(I = 2\text{ A}\)
- Time, \(t = 5\text{ minutes} = 5 \times 60 = 300\text{ s}\)
We know that \(I = \frac{Q}{t}\), therefore \(Q = I \times t\)
\(Q = 2\text{ A} \times 300\text{ s} = 600\text{ C}\)
Therefore, 600 coulombs of charge flows through the conductor in 5 minutes.
Flow of Electric Charges in a Metallic Conductor
In metallic conductors, the flow of electric charges is primarily due to the movement of free electrons. These free electrons, also known as conduction electrons, are not bound to any particular atom and can move freely through the conductor.
When a potential difference is applied across a metallic conductor, these free electrons experience an electric force and move in a direction opposite to the electric field (since electrons are negatively charged). This collective motion of electrons constitutes the electric current.
The classical free electron theory of metals explains this phenomenon well. According to this theory:
- Metals contain a large number of free electrons that behave like a “gas” of charged particles.
- These electrons move randomly in all directions with high thermal velocities (approximately 10^6 m/s at room temperature).
- When an electric field is applied, a small drift velocity (typically 10^-4 m/s) is superimposed on this random motion.
- This drift of electrons constitutes the electric current.
The current density (\(J\)) in a conductor is related to the number density of free electrons (\(n\)), the charge of an electron (\(e\)), and the drift velocity (\(v_d\)) as:
Where:
- \(J\) is the current density (A/m²)
- \(n\) is the number of free electrons per unit volume (m^-3)
- \(e\) is the elementary charge (1.6 × 10^-19 C)
- \(v_d\) is the drift velocity (m/s)
Drift Velocity
Drift velocity is the average velocity with which free electrons move in a conductor in response to an electric field. Despite the fact that individual electrons move at very high speeds (thermal velocities), their net displacement due to random motion is zero. However, when an electric field is applied, they acquire a small net velocity in the direction opposite to the electric field.
Where:
- \(v_d\) is the drift velocity (m/s)
- \(e\) is the elementary charge (1.6 × 10^-19 C)
- \(E\) is the electric field (V/m)
- \(m\) is the mass of an electron (9.1 × 10^-31 kg)
- \(\tau\) is the mean free time between collisions (s)
The current through a wire can be related to the drift velocity by:
Where:
- \(I\) is the current (A)
- \(n\) is the number of free electrons per unit volume (m^-3)
- \(e\) is the elementary charge (C)
- \(A\) is the cross-sectional area of the conductor (m²)
- \(v_d\) is the drift velocity (m/s)
Numerical Example:
A copper wire with cross-sectional area 2 × 10^-6 m² carries a current of 1 A. The number density of free electrons in copper is 8.5 × 10^28 m^-3. Calculate the drift velocity of electrons.
Solution:
Given:
- Current, \(I = 1\text{ A}\)
- Cross-sectional area, \(A = 2 \times 10^{-6}\text{ m}^2\)
- Number density of free electrons, \(n = 8.5 \times 10^{28}\text{ m}^{-3}\)
- Elementary charge, \(e = 1.6 \times 10^{-19}\text{ C}\)
Using the formula \(I = neAv_d\), we can find \(v_d\):
\(v_d = \frac{I}{neA}\)
\(v_d = \frac{1}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}\)
\(v_d = \frac{1}{2.72 \times 10^{4}}\)
\(v_d = 3.68 \times 10^{-5}\text{ m/s}\)
Therefore, the drift velocity of electrons in the copper wire is 3.68 × 10^-5 m/s or 0.0368 mm/s.
Mobility and its Relation with Electric Current
Mobility (\(\mu\)) of charge carriers in a conductor is defined as the drift velocity per unit electric field. It characterizes how quickly a charge carrier can move through a material under the influence of an electric field.
Where:
- \(\mu\) is the mobility (m²/(V·s))
- \(v_d\) is the drift velocity (m/s)
- \(E\) is the electric field (V/m)
Mobility can also be expressed in terms of the relaxation time (\(\tau\)) between collisions:
Where:
- \(e\) is the elementary charge (C)
- \(\tau\) is the mean free time between collisions (s)
- \(m\) is the mass of the charge carrier (kg)
Relation with Electric Current
Mobility is directly related to the conductivity (\(\sigma\)) of a material:
And since current density (\(J\)) is related to the electric field (\(E\)) and conductivity (\(\sigma\)) by Ohm’s law in the form \(J = \sigma E\), we can express current density in terms of mobility:
This shows that the current density (and hence the current) is directly proportional to the mobility of charge carriers. Materials with higher charge carrier mobility tend to be better conductors of electricity.
Numerical Example:
In a semiconductor, the drift velocity of electrons is 250 m/s when subjected to an electric field of 5 × 10^3 V/m. Calculate the mobility of electrons.
Solution:
Given:
- Drift velocity, \(v_d = 250\text{ m/s}\)
- Electric field, \(E = 5 \times 10^3\text{ V/m}\)
Using the formula \(\mu = \frac{v_d}{E}\):
\(\mu = \frac{250}{5 \times 10^3}\)
\(\mu = 0.05\text{ m}^2\text{/(V·s)}\)
Therefore, the mobility of electrons in the given semiconductor is 0.05 m²/(V·s).
Ohm’s Law
Ohm’s law is one of the fundamental laws in electricity. It states that the current flowing through a conductor is directly proportional to the potential difference (voltage) across it, provided the physical conditions (like temperature) remain constant.
Where:
- \(V\) is the potential difference (in volts, V)
- \(I\) is the current (in amperes, A)
- \(R\) is the resistance (in ohms, Ω)
Ohm’s law can also be written in terms of current density (\(J\)) and electric field (\(E\)):
Where \(\sigma\) is the electrical conductivity of the material.
Materials that obey Ohm’s law are called ohmic conductors, while those that don’t are called non-ohmic conductors. Most metals are ohmic conductors at normal temperatures. Examples of non-ohmic conductors include semiconductors, vacuum tubes, and some electronic components like diodes.
Numerical Example:
A resistor with a resistance of 220 Ω is connected across a 12 V battery. Calculate the current flowing through the resistor.
Solution:
Given:
- Resistance, \(R = 220\text{ Ω}\)
- Voltage, \(V = 12\text{ V}\)
Using Ohm’s law, \(V = IR\), we can find \(I\):
\(I = \frac{V}{R}\)
\(I = \frac{12}{220}\)
\(I = 0.0545\text{ A} = 54.5\text{ mA}\)
Therefore, the current flowing through the resistor is 54.5 mA.
V-I Characteristics (Linear and Non-linear)
The voltage-current (V-I) characteristic of a component is a graph that shows the relationship between the voltage across it and the current passing through it. This helps in understanding the behavior of the component under different operating conditions.
Linear (Ohmic) Characteristics
For components that obey Ohm’s law, the V-I characteristic is a straight line passing through the origin. The slope of this line represents the resistance of the component, which remains constant regardless of the applied voltage.
Examples of components with linear V-I characteristics include:
- Metallic conductors at constant temperature
- Carbon resistors
- Electrolytic solutions
Non-linear (Non-ohmic) Characteristics
Components with non-linear V-I characteristics do not follow Ohm’s law. Their resistance varies with changes in voltage or current. The V-I graph for such components is not a straight line.
Examples of components with non-linear V-I characteristics include:
- Semiconductor diodes
- Vacuum tubes
- Thermistors
- Light-dependent resistors (LDRs)
- Varistors
For these non-linear components, we often define “dynamic resistance” or “differential resistance” at a specific operating point:
This represents the slope of the V-I curve at that particular point and gives the resistance to small changes in current or voltage around that operating point.
Numerical Example:
The following data represents the voltage and current measurements for a device:
| Voltage (V) | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 |
|---|---|---|---|---|---|
| Current (mA) | 2.0 | 8.0 | 18.0 | 32.0 | 50.0 |
Determine if the device is ohmic or non-ohmic.
Solution:
To determine if the device is ohmic, we need to calculate the resistance at each point and check if it remains constant.
| Voltage (V) | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 |
|---|---|---|---|---|---|
| Current (mA) | 2.0 | 8.0 | 18.0 | 32.0 | 50.0 |
| Resistance (kΩ) | 0.5 | 0.25 | 0.167 | 0.125 | 0.1 |
Since the resistance is not constant but decreases as the voltage increases, the device is non-ohmic.
Electrical Energy and Power
When current flows through a conductor, electrical energy is converted into other forms of energy such as heat, light, or mechanical energy. The rate at which this energy is converted is called electrical power.
Electrical Energy
The electrical energy consumed by a device is the product of the power and the time for which the device operates:
Where:
- \(E\) is the electrical energy (in joules, J)
- \(P\) is the power (in watts, W)
- \(t\) is the time (in seconds, s)
In practical applications, electrical energy is often measured in kilowatt-hours (kWh):
Electrical Power
Electrical power is the rate at which electrical energy is converted into other forms of energy. It can be calculated using several equivalent formulas:
Where:
- \(P\) is the power (in watts, W)
- \(V\) is the potential difference (in volts, V)
- \(I\) is the current (in amperes, A)
- \(R\) is the resistance (in ohms, Ω)
Numerical Example 1:
An electric heater draws a current of 10 A when connected to a 220 V supply. Calculate:
a) The power of the heater
b) The resistance of the heater
c) The energy consumed in 2 hours
Solution:
Given:
- Current, \(I = 10\text{ A}\)
- Voltage, \(V = 220\text{ V}\)
- Time, \(t = 2\text{ hours} = 2 \times 60 \times 60 = 7200\text{ s}\)
a) Power of the heater:
\(P = VI = 220 \times 10 = 2200\text{ W} = 2.2\text{ kW}\)
b) Resistance of the heater:
\(R = \frac{V}{I} = \frac{220}{10} = 22\text{ Ω}\)
c) Energy consumed in 2 hours:
\(E = P \times t = 2200 \times 7200 = 15,840,000\text{ J} = 15.84\text{ MJ}\)
In kilowatt-hours: \(E = 2.2 \times 2 = 4.4\text{ kWh}\)
Numerical Example 2:
A circuit contains three resistors of 10 Ω, 15 Ω, and 20 Ω connected in series to a 12 V battery. Calculate the power dissipated in each resistor.
Solution:
Given:
- Resistances: \(R_1 = 10\text{ Ω}\), \(R_2 = 15\text{ Ω}\), \(R_3 = 20\text{ Ω}\)
- Voltage, \(V = 12\text{ V}\)
Total resistance in series:
\(R_{total} = R_1 + R_2 + R_3 = 10 + 15 + 20 = 45\text{ Ω}\)
Current in the circuit:
\(I = \frac{V}{R_{total}} = \frac{12}{45} = 0.267\text{ A}\)
Power dissipated in each resistor:
\(P_1 = I^2R_1 = (0.267)^2 \times 10 = 0.712\text{ W}\)
\(P_2 = I^2R_2 = (0.267)^2 \times 15 = 1.068\text{ W}\)
\(P_3 = I^2R_3 = (0.267)^2 \times 20 = 1.424\text{ W}\)
Total power dissipated:
\(P_{total} = P_1 + P_2 + P_3 = 0.712 + 1.068 + 1.424 = 3.204\text{ W}\)
We can verify this using \(P_{total} = VI = 12 \times 0.267 = 3.204\text{ W}\)
Electrical Resistivity and Conductivity
The resistance of a conductor depends not only on the material it is made of but also on its dimensions. Resistivity is a property of the material that is independent of the dimensions of the conductor.
Electrical Resistivity
Resistivity (\(\rho\)) is defined as the electrical resistance of a unit length and unit cross-sectional area of a material. It is a characteristic property of the material.
Where:
- \(\rho\) is the resistivity (in ohm-meter, Ω·m)
- \(R\) is the resistance (in ohms, Ω)
- \(A\) is the cross-sectional area (in m²)
- \(L\) is the length (in m)
From this, we can calculate the resistance of a conductor as:
The resistivity of a material depends on its temperature, purity, and sometimes on the applied electric field.
Electrical Conductivity
Conductivity (\(\sigma\)) is the reciprocal of resistivity. It measures the ability of a material to conduct electric current.
Where:
- \(\sigma\) is the conductivity (in siemens per meter, S/m)
- \(\rho\) is the resistivity (in ohm-meter, Ω·m)
Conductivity can also be related to the properties of charge carriers:
Where:
- \(n\) is the number density of charge carriers (in m^-3)
- \(e\) is the elementary charge (in C)
- \(\mu\) is the mobility of charge carriers (in m²/(V·s))
Numerical Example:
A copper wire has a cross-sectional area of 1.5 × 10^-6 m² and a length of 20 m. If the resistivity of copper is 1.68 × 10^-8 Ω·m, calculate:
a) The resistance of the wire
b) The conductivity of copper
Solution:
Given:
- Cross-sectional area, \(A = 1.5 \times 10^{-6}\text{ m}^2\)
- Length, \(L = 20\text{ m}\)
- Resistivity of copper, \(\rho = 1.68 \times 10^{-8}\text{ Ω·m}\)
a) Resistance of the wire:
\(R = \frac{\rho L}{A} = \frac{1.68 \times 10^{-8} \times 20}{1.5 \times 10^{-6}} = \frac{3.36 \times 10^{-7}}{1.5 \times 10^{-6}} = 0.224\text{ Ω}\)
b) Conductivity of copper:
\(\sigma = \frac{1}{\rho} = \frac{1}{1.68 \times 10^{-8}} = 5.95 \times 10^7\text{ S/m}\)
Note: Materials with high conductivity (low resistivity) are good conductors of electricity, while materials with low conductivity (high resistivity) are poor conductors or insulators. Semiconductors have intermediate values of conductivity.
| Material | Resistivity (Ω·m) at 20°C | Conductivity (S/m) at 20°C |
|---|---|---|
| Silver | 1.59 × 10^-8 | 6.29 × 10^7 |
| Copper | 1.68 × 10^-8 | 5.95 × 10^7 |
| Gold | 2.44 × 10^-8 | 4.10 × 10^7 |
| Aluminum | 2.82 × 10^-8 | 3.55 × 10^7 |
| Iron | 1.00 × 10^-7 | 1.00 × 10^7 |
| Silicon | 640 | 1.56 × 10^-3 |
| Glass | 10^10 – 10^14 | 10^-14 – 10^-10 |
Temperature Dependence of Resistance
The resistance of a conductor generally increases with temperature. This is because as temperature increases, the atoms in the conductor vibrate more vigorously, making it more difficult for the charge carriers to move through the material.
For Metals
For metallic conductors, the resistance increases almost linearly with temperature within a moderate temperature range:
Where:
- \(R_t\) is the resistance at temperature \(t\)
- \(R_0\) is the resistance at reference temperature \(t_0\) (usually 0°C or 20°C)
- \(\alpha\) is the temperature coefficient of resistance (per °C)
- \(t\) is the temperature (in °C)
- \(t_0\) is the reference temperature (in °C)
The temperature coefficient of resistance (\(\alpha\)) is a characteristic property of the material. For most metals, \(\alpha\) is positive, typically in the range of 0.003 to 0.006 per °C.
For Semiconductors
In semiconductors, the resistance generally decreases with increasing temperature. This is because more charge carriers become available as temperature increases. The relationship between resistance and temperature for semiconductors is often exponential:
Where:
- \(R_t\) is the resistance at absolute temperature \(T\)
- \(R_0\) is the resistance at reference absolute temperature \(T_0\)
- \(B\) is a constant characteristic of the material (in kelvin)
- \(T\) is the absolute temperature (in kelvin)
- \(T_0\) is the reference absolute temperature (in kelvin)
Numerical Example:
A copper wire has a resistance of 20 Ω at 20°C. If the temperature coefficient of resistance for copper is 0.00393 per °C, calculate the resistance of the wire at 80°C.
Solution:
Given:
- Resistance at 20°C, \(R_0 = 20\text{ Ω}\)
- Temperature coefficient of resistance, \(\alpha = 0.00393\text{ per °C}\)
- Reference temperature, \(t_0 = 20\text{ °C}\)
- Final temperature, \(t = 80\text{ °C}\)
Using the formula \(R_t = R_0[1 + \alpha(t – t_0)]\):
\(R_t = 20[1 + 0.00393(80 – 20)]\)
\(R_t = 20[1 + 0.00393 \times 60]\)
\(R_t = 20[1 + 0.2358]\)
\(R_t = 20 \times 1.2358\)
\(R_t = 24.72\text{ Ω}\)
Therefore, the resistance of the wire at 80°C is 24.72 Ω.
Applications: The temperature dependence of resistance is utilized in various devices such as resistance thermometers (like platinum resistance thermometers, PRTs), thermistors (temperature-sensitive resistors), and temperature sensors. Understanding this phenomenon is crucial for designing electrical systems that operate over a range of temperatures.
Internal Resistance of a Cell
An electric cell or battery is not a perfect source of voltage. It has an internal resistance that limits the current it can deliver. This internal resistance arises due to the electrolyte’s resistance, electrode-electrolyte interface resistance, and other factors.
When a current flows through a cell, part of the voltage is dropped across this internal resistance, reducing the voltage available at the terminals of the cell.
Where:
- \(V\) is the terminal voltage (the voltage measured across the terminals of the cell)
- \(\mathcal{E}\) is the electromotive force (emf) of the cell (the maximum potential difference it can provide)
- \(I\) is the current flowing through the cell
- \(r\) is the internal resistance of the cell
The terminal voltage (\(V\)) is always less than the emf (\(\mathcal{E}\)) when the cell is delivering current (discharging). When the cell is being charged, the terminal voltage is greater than the emf.
Equivalent Circuit
A real cell can be modeled as an ideal voltage source (with emf \(\mathcal{E}\)) in series with a resistor (the internal resistance \(r\)).
Determining Internal Resistance
The internal resistance of a cell can be determined by measuring the terminal voltage at different currents. If we plot terminal voltage against current, the slope of the line gives the negative of the internal resistance.
Numerical Example:
A battery with an emf of 12 V has an internal resistance of 0.5 Ω. Calculate:
a) The terminal voltage when a current of 2 A is drawn from the battery
b) The maximum current that can be drawn from the battery (short-circuit current)
Solution:
Given:
- EMF, \(\mathcal{E} = 12\text{ V}\)
- Internal resistance, \(r = 0.5\text{ Ω}\)
- Current, \(I = 2\text{ A}\) (for part a)
a) Terminal voltage when 2 A current is drawn:
Using \(V = \mathcal{E} – Ir\):
\(V = 12 – 2 \times 0.5\)
\(V = 12 – 1\)
\(V = 11\text{ V}\)
b) Maximum current (short-circuit current):
When the battery is short-circuited, the terminal voltage \(V = 0\). Using \(V = \mathcal{E} – Ir\):
\(0 = 12 – I \times 0.5\)
\(I \times 0.5 = 12\)
\(I = \frac{12}{0.5} = 24\text{ A}\)
Therefore, the short-circuit current is 24 A. However, it’s important to note that drawing such a high current can damage the battery and potentially cause safety hazards.
Note: The internal resistance of a cell increases as the cell ages or discharges. It also depends on temperature, generally increasing at lower temperatures. This is why batteries perform poorly in cold weather.
Potential Difference and EMF of a Cell
The electromotive force (emf) and potential difference are related but distinct concepts in electricity.
Potential Difference
Potential difference (voltage) is the energy required to move a unit positive charge from one point to another in an electric field. It is measured in volts (V).
Where:
- \(V\) is the potential difference (in volts, V)
- \(W\) is the work done (in joules, J)
- \(q\) is the charge (in coulombs, C)
Electromotive Force (EMF)
The electromotive force (emf) of a cell is the maximum potential difference that the cell can provide. It is the work done by non-electrostatic forces within the cell to move a unit positive charge around the complete circuit. EMF is also measured in volts (V).
The emf (\(\mathcal{E}\)) of a cell is equal to the terminal voltage (\(V\)) only when no current is flowing (open-circuit condition). When current flows, the terminal voltage is less than the emf due to the voltage drop across the internal resistance:
Sources of EMF
EMF can be generated through various processes:
- Chemical Action: In batteries and fuel cells, chemical energy is converted into electrical energy.
- Electromagnetic Induction: When a conductor moves in a magnetic field, an emf is induced (generators, alternators).
- Thermoelectric Effect: When two different metals are joined and the junctions are maintained at different temperatures, an emf is generated (thermocouples).
- Photoelectric Effect: When light falls on certain materials, they emit electrons, generating an emf (solar cells).
- Piezoelectric Effect: When mechanical stress is applied to certain crystals, they generate an emf.
Numerical Example:
A battery with an emf of 6 V and internal resistance of 0.2 Ω is connected to a resistor of 5.8 Ω. Calculate:
a) The current in the circuit
b) The terminal voltage of the battery
c) The power delivered to the external resistor
d) The power wasted in the internal resistance
e) The efficiency of the battery
Solution:
Given:
- EMF, \(\mathcal{E} = 6\text{ V}\)
- Internal resistance, \(r = 0.2\text{ Ω}\)
- External resistance, \(R = 5.8\text{ Ω}\)
a) Current in the circuit:
The total resistance in the circuit is \(R_{total} = R + r = 5.8 + 0.2 = 6\text{ Ω}\)
Using Ohm’s law: \(I = \frac{\mathcal{E}}{R_{total}} = \frac{6}{6} = 1\text{ A}\)
b) Terminal voltage of the battery:
\(V = \mathcal{E} – Ir = 6 – 1 \times 0.2 = 5.8\text{ V}\)
c) Power delivered to the external resistor:
\(P_R = I^2R = 1^2 \times 5.8 = 5.8\text{ W}\)
d) Power wasted in the internal resistance:
\(P_r = I^2r = 1^2 \times 0.2 = 0.2\text{ W}\)
e) Efficiency of the battery:
\(\eta = \frac{P_R}{P_R + P_r} \times 100\% = \frac{5.8}{5.8 + 0.2} \times 100\% = \frac{5.8}{6} \times 100\% = 96.67\%\)
Alternatively, the efficiency can be calculated as:
\(\eta = \frac{V}{\mathcal{E}} \times 100\% = \frac{5.8}{6} \times 100\% = 96.67\%\)
Kirchhoff’s Rules
Kirchhoff’s rules are two fundamental laws that deal with the conservation of current and energy in electrical circuits. They are particularly useful for analyzing complex circuits with multiple loops and branches.
Kirchhoff’s Current Law (KCL)
This law is based on the principle of conservation of charge. It states that at any junction (node) in an electrical circuit, the sum of currents flowing into the node is equal to the sum of currents flowing out of the node.
Kirchhoff’s Voltage Law (KVL)
This law is based on the principle of conservation of energy. It states that in any closed loop in a circuit, the sum of all voltage rises (emfs) equals the sum of all voltage drops (across resistors, etc.).
When applying KVL, we need to choose a direction to traverse the loop. By convention:
- If we traverse from the negative to the positive terminal of a battery, it is a voltage rise (+\(\mathcal{E}\)).
- If we traverse from the positive to the negative terminal, it is a voltage drop (−\(\mathcal{E}\)).
- If we traverse a resistor in the direction of current, it is a voltage drop (−\(IR\)).
- If we traverse a resistor against the direction of current, it is a voltage rise (+\(IR\)).
Numerical Example:
Consider the following circuit:
Two batteries with emfs 12 V and 6 V and internal resistances 1 Ω and 0.5 Ω respectively are connected in the same loop with a resistor of 20 Ω. The positive terminal of the 12 V battery is connected to the negative terminal of the 6 V battery. Find the current in the circuit.
Solution:
Let’s apply Kirchhoff’s voltage law to this circuit. We’ll assume the current \(I\) flows from the positive terminal of the 12 V battery, through the 20 Ω resistor, and back to its negative terminal.
Going around the loop, we have:
\(12 – I \times 1 – I \times 20 + 6 – I \times 0.5 = 0\)
\(12 + 6 – I(1 + 20 + 0.5) = 0\)
\(18 – 21.5I = 0\)
\(I = \frac{18}{21.5} = 0.837\text{ A}\)
Therefore, the current in the circuit is 0.837 A, flowing from the positive terminal of the 12 V battery through the 20 Ω resistor and back to its negative terminal.
Numerical Example (Complex Circuit):
Consider a circuit with two loops. In the first loop, a 10 V battery with internal resistance 1 Ω is connected to a 5 Ω resistor. In the second loop, a 5 V battery with internal resistance 0.5 Ω is connected to a 4.5 Ω resistor. The two loops share a common 2 Ω resistor. Find the currents in each branch of the circuit.
Solution:
Let’s call the current in the first loop \(I_1\) and the current in the second loop \(I_2\). The current through the common resistor will be \(I_1 – I_2\) (assuming \(I_1\) and \(I_2\) are in opposite directions through the common resistor).
Applying KVL to the first loop:
\(10 – I_1 \times 1 – I_1 \times 5 – (I_1 – I_2) \times 2 = 0\)
\(10 – I_1 – 5I_1 – 2I_1 + 2I_2 = 0\)
\(10 – 8I_1 + 2I_2 = 0\) … (1)
Applying KVL to the second loop:
\(5 – I_2 \times 0.5 – I_2 \times 4.5 + (I_1 – I_2) \times 2 = 0\)
\(5 – 0.5I_2 – 4.5I_2 + 2I_1 – 2I_2 = 0\)
\(5 – 7I_2 + 2I_1 = 0\) … (2)
From equation (1): \(8I_1 – 2I_2 = 10\) … (3)
From equation (2): \(-2I_1 + 7I_2 = 5\) … (4)
Multiplying equation (3) by 7 and equation (4) by 8:
\(56I_1 – 14I_2 = 70\) … (5)
\(-16I_1 + 56I_2 = 40\) … (6)
Adding equations (5) and (6):
\(40I_1 + 42I_2 = 110\)
\(42I_2 = 110 – 40I_1\) … (7)
From equation (3): \(2I_2 = 8I_1 – 10\) … (8)
Multiplying equation (8) by 21:
\(42I_2 = 168I_1 – 210\) … (9)
Equating (7) and (9):
\(110 – 40I_1 = 168I_1 – 210\)
\(110 + 210 = 40I_1 + 168I_1\)
\(320 = 208I_1\)
\(I_1 = \frac{320}{208} = 1.54\text{ A}\)
Substituting back into equation (8):
\(2I_2 = 8 \times 1.54 – 10\)
\(2I_2 = 12.32 – 10\)
\(2I_2 = 2.32\)
\(I_2 = 1.16\text{ A}\)
Therefore, the current in the first loop is 1.54 A and the current in the second loop is 1.16 A. The current through the common resistor is \(I_1 – I_2 = 1.54 – 1.16 = 0.38\text{ A}\).
Wheatstone Bridge
The Wheatstone bridge is a circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component.
Construction
A Wheatstone bridge consists of four resistors arranged in a diamond shape. A voltage source is connected between two opposite corners of the diamond, and a galvanometer or other detector is connected between the other two corners.
Let’s denote the four resistors as \(R_1\), \(R_2\), \(R_3\), and \(R_4\). The voltage source is connected between the junction of \(R_1\) and \(R_4\) and the junction of \(R_2\) and \(R_3\). The galvanometer is connected between the junction of \(R_1\) and \(R_2\) and the junction of \(R_3\) and \(R_4\).
Balanced Condition
The bridge is said to be balanced when there is no current flowing through the galvanometer. This occurs when the potential difference across the galvanometer is zero, which happens when the ratios of the resistances in the two legs are equal:
This relationship can be rearranged to find an unknown resistance. For example, if \(R_4\) is the unknown resistance, we can compute it as:
Applications
The Wheatstone bridge has several applications:
- Precision Resistance Measurement: By using known precise resistors for \(R_1\), \(R_2\), and \(R_3\), an unknown resistance \(R_4\) can be measured with high accuracy.
- Strain Gauge Measurements: Strain gauges often use Wheatstone bridges to measure small changes in resistance due to strain.
- Temperature Measurement: Using temperature-sensitive resistors (thermistors), the bridge can be used to measure temperature.
- Pressure Measurement: Pressure-sensitive resistors can be incorporated into a Wheatstone bridge for precise pressure measurements.
Numerical Example:
In a Wheatstone bridge, three resistors have values \(R_1 = 100\text{ Ω}\), \(R_2 = 200\text{ Ω}\), and \(R_3 = 300\text{ Ω}\). If the bridge is balanced, find the value of the fourth resistor \(R_4\).
Solution:
Given:
- \(R_1 = 100\text{ Ω}\)
- \(R_2 = 200\text{ Ω}\)
- \(R_3 = 300\text{ Ω}\)
When the bridge is balanced, we have:
\(\frac{R_1}{R_2} = \frac{R_3}{R_4}\)
Rearranging to find \(R_4\):
\(R_4 = \frac{R_2 \times R_3}{R_1}\)
\(R_4 = \frac{200 \times 300}{100}\)
\(R_4 = \frac{60000}{100}\)
\(R_4 = 600\text{ Ω}\)
Therefore, the value of the fourth resistor is 600 Ω.
Numerical Example (Unbalanced Bridge):
In a Wheatstone bridge, the resistors have values \(R_1 = 100\text{ Ω}\), \(R_2 = 200\text{ Ω}\), \(R_3 = 250\text{ Ω}\), and \(R_4 = 450\text{ Ω}\). A battery of 10 V is connected across the bridge, and a galvanometer with resistance 25 Ω is connected between the junctions. Calculate the current through the galvanometer.
Solution:
This is a complex problem that requires us to analyze the unbalanced bridge using Kirchhoff’s laws. I’ll provide a simplified approach using equivalent resistance calculations.
First, let’s check if the bridge is balanced:
\(\frac{R_1}{R_2} = \frac{100}{200} = 0.5\)
\(\frac{R_3}{R_4} = \frac{250}{450} = 0.556\)
Since these ratios are not equal, the bridge is unbalanced and there will be current through the galvanometer.
The voltage across the galvanometer (between points B and D) can be calculated using voltage division formulas. The voltages at points B and D with respect to point A are:
\(V_B = 10 \times \frac{R_2}{R_1 + R_2} = 10 \times \frac{200}{100 + 200} = 10 \times \frac{200}{300} = 6.67\text{ V}\)
\(V_D = 10 \times \frac{R_4}{R_3 + R_4} = 10 \times \frac{450}{250 + 450} = 10 \times \frac{450}{700} = 6.43\text{ V}\)
The potential difference across the galvanometer is:
\(V_{BD} = V_B – V_D = 6.67 – 6.43 = 0.24\text{ V}\)
Now, to find the current through the galvanometer, we need to consider the equivalent resistance of the entire circuit. This is a complex calculation, but a simplified approach is to use Thévenin’s theorem.
The Thévenin equivalent voltage across BD is 0.24 V, as calculated above.
The Thévenin equivalent resistance seen from the galvanometer terminals is:
\(R_{Th} = \frac{R_1 \times R_2}{R_1 + R_2} + \frac{R_3 \times R_4}{R_3 + R_4}\)
\(R_{Th} = \frac{100 \times 200}{100 + 200} + \frac{250 \times 450}{250 + 450}\)
\(R_{Th} = \frac{20000}{300} + \frac{112500}{700}\)
\(R_{Th} = 66.67 + 160.71 = 227.38\text{ Ω}\)
The current through the galvanometer is:
\(I_G = \frac{V_{Th}}{R_{Th} + R_G} = \frac{0.24}{227.38 + 25} = \frac{0.24}{252.38} = 0.00095\text{ A} = 0.95\text{ mA}\)
Therefore, the current through the galvanometer is approximately 0.95 mA.
Note: The Wheatstone bridge is most sensitive when all four resistors have similar values. For maximum sensitivity, the galvanometer should have the highest possible sensitivity (lowest current required for deflection) and the battery should provide a stable voltage source.
