Definition: The flow of electric charge through a conductor.
Unit: Ampere (A)
Direction: Conventionally taken as the direction of flow of positive charge (opposite to the actual flow of electrons).
Ohm’s Law:
Relates voltage (V), current (I), and resistance (R) in a circuit: V = IR.
Resistance: Property of a conductor that opposes the flow of electric current. Unit: Ohm (Ω).
Resistivity: Material property that determines resistance.
Electric Circuits:
Series Combination: Resistances are connected end-to-end. Equivalent resistance (R_eq) is the sum of individual resistances: R_eq = R₁ + R₂ + …
Parallel Combination: Resistances are connected across common points. Equivalent resistance is given by: 1/R_eq = 1/R₁ + 1/R₂ + …
Kirchhoff’s Laws:
Junction Rule: The algebraic sum of currents at a junction is zero.
Loop Rule: The algebraic sum of potential differences in a closed loop is zero.
Electrical Power and Energy:
Electric Power: The rate at which electrical energy is converted into other forms of energy in an electric circuit..
It is calculated as P = VI = I²R
= V²/R.
Electric Energy: Total energy consumed. E = Pt.
Heating Effect of Electric Current:
Joule’s Law: Heat produced (H) = I²Rt.
Applications: Electric heaters, geysers, toasters, etc.
Cells
Primary Cells: Cannot be recharged (e.g., dry cells).
Secondary Cells: Can be recharged (e.g., lead-acid batteries).
Potentiometer:
A device to measure potential difference.
Used to compare potential differences, measure emf of a cell, and determine internal resistance of a cell.
This chapter provides a foundation for understanding the fundamentals of electricity and its applications in various devices.
1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
Ans : We can use Ohm’s Law to determine the maximum current.
Ohm’s Law: V = IR
In this case, the battery’s internal resistance (R) is 0.4 Ω, and its electromotive force (emf) is 12 V.
To maximize the current, we want to minimize the voltage drop across the internal resistance. This occurs when the external resistance is zero (short circuit).
So, the maximum current (I_max) can be calculated as:
I_max = V / R = 12 V / 0.4 Ω
= 30 A
Therefore, the maximum current that can be drawn from the battery is 30 amperes. However, it’s important to note that drawing such a high current for an extended period can damage the battery.
2. A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Ans :Understanding the Problem:**
We have a battery with a given emf and internal resistance connected to an external resistor. We know the current flowing through the circuit and need to find the external resistance and the terminal voltage of the battery.
1. Finding the External Resistance:
We can use Ohm’s law to find the total resistance in the circuit:
V = IR
Where:
V is the total voltage (emf of the battery)
I is the current
R is the total resistance
Substituting the values:
10 V = 0.5 A × R
Solving for R:
R = 10 V
————
0.5 A
= 20 Ω
This total resistance is the sum of the internal resistance (r) and the external resistance (R_external):
R = r + R_external
So, the external resistance is:
R_external = R – r = 20 Ω – 3 Ω
= 17 Ω
2. Finding the Terminal Voltage:
Terminal voltage equals the voltage drop across the external resistor.. It can be calculated using Ohm’s law again:
V_terminal = IR_external = 0.5 A × 17 Ω
= 8.5 V
Therefore, the resistance of the resistor is 17 Ω, and the terminal voltage of the battery when the circuit is closed is 8.5 V.
3. At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1 .
Ans : Given Data:
Initial temperature, T = 27°C
Initial resistance, R = 100 Ω
Final resistance, R₁ = 117 Ω
Temperature coefficient,
α = 1.70 × 10⁻⁴ °C⁻¹
Formula for Temperature Coefficient:
α = (R₁ – R) / R(T₁ – T)
Rearranging the Formula:
T₁ – T = (R₁ – R) / Rα
Substituting Values:
T₁ – 27 = (117 – 100) / (100 × 1.70 × 10⁻⁴)
Calculating Temperature Increase:
T₁ – 27 = 1000
T₁ = 1027°C
Consequently, the heating element’s resistance increases to 117 Ω when its temperature reaches 1027°C.
4. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2 , and its resistance is measured to be 5.0 W. What is the resistivity of the material at the temperature of the experiment?
Ans :Understanding the Problem:**
We are given the length, cross-sectional area, and resistance of a wire. We need to find its resistivity.
The resistance (R) of a wire is related to its resistivity (ρ), length (L), and cross-sectional area (A) by the formula:
R = ρL/A
By rearranging the formula, we can determine the resistivity of the material.
ρ = RA/L
Substituting the given values:
ρ = (5.0 Ω)(6.0 × 10⁻⁷ m²) / 15 m
= 2.0 × 10⁻⁷ Ωm
Therefore, the resistivity of the material at the temperature of the experiment is 2.0 × 10⁻⁷ Ωm.
5. A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver.
Ans :Given Data:
Initial temperature,
T₁ = 27.5°C
Initial resistance,
R₁ = 2.1 Ω
Final temperature,
T₂ = 100°C
Final resistance,
R₂ = 2.7 Ω
Formula for Temperature Coefficient:
α = (R₂ – R₁)
—————
R₁(T₂ – T₁)
Substituting Values:
α = (2.7 – 2.1)
————————————-
(2.1 × (100 – 27.5))
= 0.0039 °C⁻¹
Hence, the temperature coefficient of resistance for silver approximately 0.0039 °C⁻¹.
6. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1 .
Ans :Initial Resistance:
R₁ = V / I₁ = 230 V / 3.2 A = 71.87 Ω
Steady-State Resistance:
R₂ = V
——–
I₂
= 230 V / 2.8 A
= 82.14 Ω
Using the Temperature Coefficient Formula:
α =
(R₂ – R₁)
————–
(R₁(T₂ – T₁))
Solving for T₂:
T₂ – 27°C = (82.14 – 71.87) / (71.87 × 1.70 × 10⁻⁴)
T₂ = 840.5 + 27
= 867.5°C
Therefore, the steady-state temperature of the heating element is 867.5°C.
7. Determine the current in each branch of the network shown in Fig. 3.20:
Ans :We can apply Kirchhoff’s laws to determine the current flowing through each branch of the network.
Kirchhoff’s Current Law (KCL): The algebraic sum of currents at a junction is zero.
Kirchhoff’s Voltage Law (KVL): The algebraic sum of potential differences in a closed loop is zero.
Steps to solve:
Assign currents to each branch: Let’s assign currents I₁, I₂, and I₃ to the branches as shown in the figure below:
circuit with currents I₁, I₂, and I₃ assigned
Apply KCL at junction A:
I₁ + I₂ = I₃
Apply KVL to the outer loop:
10V – 10Ω * I₁ – 5Ω * I₃ = 0
Apply KVL to the inner loop:
-5Ω * I₃ + 10Ω * I₂ – 10V = 0
We now have a system of three equations with three unknown variables: I₁, I₂, and I₃.. We can solve these equations simultaneously to find the currents in each branch.
Solving the equations:
From equation (1), we get:
I₁ = I₃ – I₂
Putting this into equations (2) and (3), we obtain:
10 – 10(I₃ – I₂) – 5I₃ = 0
-5I₃ + 10I₂ – 10 = 0
Simplifying these equations:
-15I₃ + 10I₂ = -10
-5I₃ + 10I₂ = 10
Solving these equations simultaneously, we get:
I₃ = -1 A
I₂ = 0 A
Substituting I₃ back into equation (1), we get:
I₁ = -1 A – 0 A = -1 A
Interpretation of the negative signs:
The negative values of I₁ and I₃ imply that the actual direction of the current flow is opposite to the direction we initially assumed.
Therefore, the current in the 10Ω resistor connected to the 10V battery is 1 A, flowing from the positive terminal of the battery towards the junction A. The current in the 5Ω resistor connected between junctions A and B is 0 A. And the current in the 5Ω resistor connected between junctions B and C is 1 A, flowing from junction B towards junction C.
8 .A storage battery of emf 8.0 V and internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistor of 15.5 W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Ans :To determine the terminal voltage and the role of the series resistor, we must first calculate the current flowing through the circuit.
Calculating the Current:
The circuit’s total resistance is the combined resistance of the battery’s internal resistance and the series resistor.
Total resistance (R) = 0.5 Ω + 15.5 Ω = 16 Ω
The current (I) flowing through the circuit, according to Ohm’s law, is:
I = V
——
R
= 120 V / 16 Ω
= 7.5 A
Calculating the Terminal Voltage:
given,
(V_terminal) denote voltage measured across the battery’s terminals while charging
V_terminal = emf + Ir
Where:
emf denote battery’s electromotive force (8.0 V)
I is the current (7.5 A)
r is the internal resistance (0.5 Ω)
Substituting the values:
V_terminal = 8.0 V + (7.5 A * 0.5 Ω) = 11.75 V
Purpose of the Series Resistor:
The series resistor serves two main purposes:
Current Limiting: It restricts the current flowing through the battery during charging. Excessive charging current can damage the battery. The resistor ensures a safe charging rate.
Voltage Dropping: It reduces the voltage from the supply, ensuring the battery’s terminal voltage is slightly higher than its emf. This is crucial for forcing current into the battery and initiating the charging process.
In summary:
The terminal voltage during charging is 11.75 V.
The series resistor limits the charging current and helps maintain a voltage slightly higher than the battery’s emf, facilitating the charging process.
9 .The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A.
Ans :To determine the time it takes for an electron to drift from one end of a 3-meter copper wire to the other, we need to consider the following factors:
(n) denote Number density of free electrons =
8.5 x 10^28 electrons per cubic meter
(l) denote Length of the wire =
3.0 meters
(A) denote Cross-sectional area of the wire :=
2.0 x 10^-6 square meters
(I) denote Current flowing through the wire :=
3.0 amperes
(e) denote Charge of an electron : =
1.6 x 10^-19 coulombs
Step 1: Calculate Drift Velocity (v_d)
Drift velocity is the average speed of electrons in the wire. We can calculate it using the formula:
v_d =
I
———————
(n * e * A)
Substituting the given values:
v_d =
3.0 A
—————————————————————————————
(8.5 x 10^28 m^-3 * 1.6 x 10^-19 C * 2.0 x 10^-6 m^2)
This gives us:
v_d ≈ 1.10 x 10^-4 meters per second
Step 2: Calculate Drift Time (t)
The time it takes for an electron to move from one end of the wire to the other is called the drift time.
Determine this time using by following formula:
t = l
———————
v_d
Substituting the values:
t =
3.0 m
————————–
1.10 x 10^-4 m/s
This gives us:
t ≈ 2.73 x 10^4 seconds
Therefore, an electron takes approximately 2.73 x 10^4 seconds, or about 7.6 hours, to drift from one end of the wire to the other.
