৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি [২৮১]

A voltmeter is connected in parallel across two points in a circuit to measure the potential difference (voltage) between them. It ideally has very high resistance to prevent current from flowing through it.

Example:
To measure the voltage across a resistor in a circuit, a voltmeter is connected across the resistor terminals.

Sources: Fundamentals of Physics by Halliday, Resnick & Walker

Ohm’s Law states:
V = IR

Where,
V = voltage across the resistor
I = current through the resistor
R = resistance

If R is constant and V is doubled, the current I = V/R doubles.

Sources: Fundamentals of Physics by Halliday, Resnick & Walker

Voltage (also called electric potential difference) measures the energy per unit charge that is required to move a charge between two points in an electric field.

Mathematically:  V = W/Q

Where,
V = Voltage (in volts)
W = Work done or energy (in joules)
Q = Electric charge (in coulombs)

1 Volt = 1 Joule / 1 Coulomb

Sources: Fundamentals of Physics by Halliday, Resnick & Walker

Ohm’s Law defines the relationship between voltage (V), current (I), and resistance (R) in a circuit. The law states that the voltage across a resistor is directly proportional to the current flowing through it and the resistance of the resistor:
V=IR

Where,
V is the voltage (in volts, V),
I is the current (in amperes, A),
R is the resistance (in ohms, Ω).

This means that if the current increases, the voltage will also increase (for a constant resistance), and if the resistance increases, the voltage across the resistor will increase (for a constant current).

Sources: Fundamentals of Physics by Halliday, Resnick & Walker

The SI unit of electric current is the Ampere (A). The ampere is one of the seven base units in the International System of Units (SI). It is defined as the flow of one coulomb of charge per second.

Definition of Ampere (A):
1 Ampere (A) is the current that flows when 1 coulomb of charge passes through a conductor in 1 second.

Mathematically:
I = Q/t
 ​
Where,
I is the current in amperes (A),
Q is the charge in coulombs (C),
t is the time in seconds (s).


Sources: Fundamentals of Physics by Halliday, Resnick & Walker

The efficiency of an electrical appliance is the ratio of the useful power output to the total power input, and it is usually expressed as a percentage.
This represents how effectively the appliance converts input energy into useful work (power).
     
Source: Electrical Engineering: Principles and Applications by Allan R. Hambley

The SI unit of energy is the joule (J), which is defined as the energy required to move one coulomb of charge through one volt of potential difference. It is also the energy used when a force of one newton acts over a distance of one meter.
1J=1N⋅m

Example:
If you lift a 1 kg object by 1 meter, you use approximately 9.8 joules of energy.

Source: Physics for Scientists and Engineers by Serway and Jewett

Kinetic Energy, Potential Energy, and Thermal Energy are all forms of energy. Voltage is the potential difference, which is related to the energy per unit charge but is not a form of energy itself.

Example:
A 12 V battery → voltage (ability to supply energy, but not energy itself)

Source: Physics for Scientists and Engineers by Serway and Jewett

According to the work-energy theorem, the work done on an object results in a change in its energy, specifically kinetic energy.
Mathematically, the work done (W) equals the change in kinetic energy (ΔK).

Example:
If a car accelerates from rest to a higher speed, the work done on it changes its kinetic energy.


Source: Physics for Scientists and Engineers by Serway and Jewett

Potential Energy is the energy stored in an object due to its position or configuration. A compressed spring has potential energy because it can do work when it is released.

Example:
A stretched rubber band or a raised object also has potential energy, which can be converted to kinetic energy.

Source: Physics for Scientists and Engineers by Serway and Jewett

The formula to calculate the gravitational potential energy (E_P) is:
E_P =mgh

Where
m is the mass of the object (in kilograms),
g is the acceleration due to gravity (9.8 m/s² ),
h is the height from the ground (in meters).

Given that,
Mass (m) = 5 kg,
Height (h) = 10 m,
Gravitational acceleration (g) =9.8 m/s²

Now, substituting the values into the formula,

E_P =5×9.8 ×10
   = 490

E_P = 490 Joules

∴ The gravitational potential energy of the object is 490 Joules (J).

Source: Physics for Scientists and Engineers by Serway and Jewett

A dependent source (also called a controlled source) produces an output that depends on another voltage or current within the same circuit. It is usually used to model devices like transistors or operational amplifiers.

Example:
Voltage-controlled voltage source (VCVS):
The output voltage is a fixed multiple of the controlling input voltage (e.g., V_out =2×V_control ).

Current-controlled current source (CCCS):
The output current is a fixed multiple of the controlling input current (e.g., I_out = 3×I_control ).

Source: Physics for Scientists and Engineers by Serway and Jewett

An independent current source is not dependent on any other variable in the circuit. It provides a fixed current regardless of other circuit parameters. All other options are examples of dependent sources, where the output is controlled by a voltage or current elsewhere in the circuit.

Example:
Independent current source: A current source that provides a constant current (e.g., 2A) regardless of the voltage.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Resistance is the property of a material that opposes the flow of electric current. The resistance of a material is determined by its resistivity. Among the given options, rubber has the highest resistance because it is an insulator and resists current flow very effectively. Metals like copper, aluminum, and silver are good conductors and have low resistance.

Example:
Copper and silver are commonly used in electrical wiring due to their low resistance, while rubber is used as insulation because of its high resistance.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a series circuit, the total resistance (R) is the sum of the individual resistances:

R = R₁+ R₂ + R₃
So, for the given resistors, R = 4Ω+6Ω+10Ω=20Ω

If you connect three resistors of 4Ω, 6Ω, and 10Ω in series, the total resistance is 20Ω.
The right answer is 20Ω

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

To calculate the current flowing through the resistor, we use Ohm's Law, which states,
I = V/R

Where,
I is the current in amperes (A),
V is the voltage across the resistor in volts (V),
R is the resistance in ohms (Ω).

Given that,
Voltage, V = 12V,
Resistance, R = 6Ω.

Now, substituting the values into the formula,
I = V/R
 = 12/6
 = 2
∴ I = 2 A

So, the current flowing through the resistor is 2A.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

According to Ohm’s Law, I = V/R, if the voltage remains constant and the resistance is doubled, the current will be halved. This is because current is inversely proportional to resistance.

Example:
If the voltage is constant at 10V and the resistance is 5Ω, the current will be 2A. If the resistance is doubled to 10Ω, the current will be 1A, which is half the previous current.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Ohm’s Law is most accurate when the temperature of the conductor is constant because temperature variations can change the resistance of the conductor. For most materials (like metals), resistance increases with temperature, which can deviate from Ohm's Law. Ohm’s Law assumes the resistance remains constant.

Example:
A metal wire at room temperature follows Ohm’s Law well. However, if the wire is heated, its resistance increases, which would make the current deviate from Ohm’s Law if the voltage remains constant.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

According to Ohm’s Law, I = V/R. If the voltage remains constant and the resistance decreases, the current increases. This is because current is inversely proportional to resistance.

Example:
If the voltage is 12V and the resistance is 6Ω, the current is I = 12/6 =2A. If the resistance is reduced to 3Ω, the current increases to I =12/3 = 4A.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Kirchhoff’s Current Law states,
“The algebraic sum of currents entering a junction (or node) in an electrical circuit is equal to the sum of currents leaving the junction.”

Mathematically:
∑I_in = ∑I_out
∴ I₁+I₂ = I₃

Now, I₃ = 2A+3A =5A
The right answer is 5A

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Kirchhoff’s Current Law states,
“The algebraic sum of currents entering a junction (or node) in an electrical circuit is equal to the sum of currents leaving the junction.”

KCL comes from the conservation of electric charge, meaning charge cannot accumulate at a node; whatever enters must leave.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

We are asked to find the voltage drop across R₁ . In a series circuit, the voltage drop across a resistor is given by Ohm's Law,
V=I⋅R

Where:
V is the voltage drop across the resistor, V = IR₁
I is the current flowing through the resistor, I = 2A
R is the resistance of the resistor.

Given that,
I= 2 A
R = 2Ω

Here,
V_R1 = I⋅R₁
 ​      = 2A×2Ω=4V

∴ The right answer is 4V.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

n a series circuit, the current is the same throughout all components because there is only one path for the current to flow. This means that the same current flows through all resistors in the circuit, regardless of the number of resistors.

In a series circuit, the current has only one path to take, and it flows through every component one after another.
The total resistance in a series circuit is the sum of the individual resistances: 
R_total =R₁+R₂+R₃+…
Since there's only one path for the current, the current that flows through each resistor is the same.

Example:
A 5Ω and 10Ω resistor in series with 3A current → same 3A flows through both.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In parallel, reciprocal of total resistance is sum of reciprocals.
1/R = 1/R₁ + 1/R₂ 

Given that,
Resistance, R₁=4Ω and R₂= 6Ω 
Total resistance, R =?

Solution here, 
1/R = 1/4 + 1/6
      = 5/12
∴R = 12/5 = 2.4Ω 
The right answer is 2.4Ω 

Example:
Total current from battery:
I = V/R
 = 12/2.4
 = 5
I = 5A

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a series circuit, the voltage across each resistor is proportional to its resistance. The voltage division rule for a series circuit is given by:

V_x = V × ( R_x / R)

Where,
V is the total voltage across the series circuit (Given that 18 V),
R_x is the resistance of the resistor of interest,
R is the total resistance of the circuit.

Total resistance in the series circuit:
R = R₁+R₂+R₃
   = 3Ω+6Ω+9Ω
   = 18Ω

Voltage across R₂ (6Ω):
So, V₂ = V × (R_x / R)
          = 18 × (6/18)
         = 6
∴ V₂ = 6V

 The voltage drop across R₂ is 6V.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a parallel circuit, adding more resistors increases the total resistance (since the total resistance in parallel is always lower than the smallest resistance in the circuit). As resistance increases, the total current decreases for a given voltage.

Example:
If two resistors R₁​=4Ω and R₂=8Ω are connected in parallel with a 12V battery, the total current is determined by the total resistance, which is less than either resistor. If you add a third resistor with a higher resistance, the total current will decrease.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Using the formula for source transformation:
I = V/R = 24V/12Ω = 2A
Place this current source in parallel with the same 12Ω resistor.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Source transformation is a technique used to simplify circuit analysis by converting between equivalent forms of sources. The transformation process involves changing:
•A series voltage source with a series resistor to an equivalent parallel current source with a parallel resistor.
•A parallel current source with a parallel resistor to an equivalent series voltage source with a series resistor.

Source transformation does not change the power delivered to the load. The total power delivered remains the same, as the transformation preserves the relationship between the voltage, current, and resistance in the circuit. Therefore, the power delivered to the load is unchanged, even though the source forms may be altered. This is because:
P=V×I
When transforming sources, the voltage and current adjust accordingly to maintain the same power.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Nodal analysis involves solving for the voltage at each node, which can be easier in circuits with fewer components and where there are many current sources. Mesh analysis, on the other hand, is often easier when the circuit contains more voltage sources, as it directly solves for mesh currents.

Example:
In a circuit with multiple nodes and fewer components, solving for voltages using KCL (nodal analysis) is typically simpler.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

When applying KCL in nodal analysis, the current leaving a node is considered negative. This is because KCL states that the sum of currents entering and leaving a node must be zero.

Example:
If there are currents I₁,I₂ and I₃ entering a node, and current I₄ leaving the node, the KCL equation would be:

I₁₊I₂+I₇-I_{4 =0}

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Nodal analysis focuses on calculating the voltages at each node in a circuit. From these voltages, other quantities such as currents through components can be determined using Ohm’s Law.

Example:
In a circuit with three nodes and resistors, the voltage at each node is found by solving the system of equations derived from Kirchhoff’s Current Law (KCL), and then currents can be calculated.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Thevenin’s theorem states that any linear electrical network with voltage sources, current sources, and resistors can be replaced by an equivalent voltage source (V_th ) in series with an equivalent resistance (R_th ).

This simplification makes it easier to analyze the circuit, particularly for finding the current or voltage at a specific point in the circuit.

Example:
For a circuit with a voltage source V=12V and resistors R₁=4Ω, R₂=6Ω, Thevenin’s theorem would reduce the network to a single 12V source in series with an equivalent resistance.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

To find the Norton current I_N, calculate the current through the load resistor when it is short-circuited (i.e., the load resistor is replaced by a wire). This current is the Norton current.
The short-circuit current reflects the current supplied by the equivalent current source.

Example:
If a resistor is short-circuited and a current of 3A is measured, the Norton current I_N=3A.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The superposition theorem states that in a linear circuit with multiple independent sources (voltage and current sources), the total response (current or voltage) can be found by:
  1. Considering one independent source at a time.
  2. Replacing all other independent sources by their zero-value equivalents (voltage sources replaced by short circuits and current sources replaced by open circuits)
  3. Calculating the response due to each source.
  4. Adding the responses due to all sources to get the total response.

Example:
For a circuit with two voltage sources, you calculate the current due to one source while replacing the other with a short circuit. Then, you repeat for the other source and add the currents.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

When applying the superposition theorem, each current source is replaced by an open circuit when you are considering the effect of a voltage source. Similarly, when considering a current source, all voltage sources are replaced by short circuits.

Example:
In a circuit with two voltage sources and one current source, to analyze the current due to one voltage source, replace the current source by an open circuit.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The Maximum Power Transfer Theorem states that the maximum power is transferred to the load when the load resistance R_L is equal to the Thevenin resistance R_th of the source. In this condition, the power delivered to the load is maximized.

Mathematical Expression:

Where,
P_max = The Maximum Power
V_th = The Thevenin voltage,
R_th = The Thevenin resistance.

Example:
If the Thevenin resistance of a network is 10Ω, the maximum power will be delivered to the load when the load resistance is also 10Ω.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

When the load resistance becomes much larger than the Thevenin resistance, the current flowing through the circuit decreases, resulting in a decrease in the power delivered to the load. This is because power is directly proportional to the square of the current, and as the current decreases, the power also decreases.

Example:
If R_L becomes much larger than R_th, the current drops significantly, and so does the power.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

According to the reciprocity theorem, if the positions of the current source and the measurement point are swapped, the voltage produced at the new measurement point remains the same as it was at the original measurement point.

Example:
A current source at point A generates a voltage at point B. If the source is moved to point B, the same voltage will be observed at point A due to the reciprocity theorem.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The unit of inductance is the Henry (H). One henry is the inductance of a coil in which a current change of one ampere per second induces a voltage of one volt.

Example:
An inductor of 1H would induce a 1V voltage when the current through it changes at the rate of 1A/s.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

An underdamped series RLC circuit is characterized by the following:
•The system exhibits oscillatory behavior due to the inductive and capacitive elements.
•However, the oscillations gradually decay over time because of the resistance present in the circuit. The resistance dissipates energy, leading to a decrease in the amplitude of the oscillations.
•The current in the circuit does not remain constant or increase over time. Instead, it oscillates while gradually decaying.

Key point of RLC:
Underdamped means that the damping is not strong enough to completely stop the oscillations, but strong enough to prevent them from growing unbounded.
The current initially oscillates because of the energy exchange between the inductor (L) and capacitor (C), but as time progresses, the resistance (R) causes the oscillations to dampen (decay) over time.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

In a critically damped circuit, the current decays to zero as quickly as possible without oscillating. This occurs when the damping factor ζ=1, meaning the circuit is in an ideal state for the fastest return to equilibrium without overshoot.

Example:
A critically damped RLC circuit will not oscillate, and the current will decay in the fastest possible time, without any overshoot or oscillation.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

The Wye to Delta conversion is used to simplify the analysis of circuits that have Wye-connected loads, especially when solving for current and voltage. In some cases, it can make it easier to find the equivalent resistance of the circuit.

Example:
In a complex circuit with Wye-connected resistors, converting to Delta form simplifies the circuit by reducing the number of nodes and simplifying current division.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The primary use of Wye-Delta and Delta-Wye conversions is to simplify complex resistive circuits, especially when dealing with combinations of resistors in series and parallel. The conversion makes the circuit easier to solve by reducing the number of resistors and simplifying the current and voltage analysis.

Example:
In circuits with complex resistor networks, converting between Wye and Delta configurations can make it easier to find the total equivalent resistance, current, and voltage at specific points.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The Maximum Power Transfer Theorem states that the load resistance must be equal to the Thevenin resistance of the source for maximum power to be delivered to the load. This ensures the highest possible power transfer from the source to the load.

Example:
In a circuit with a Thevenin equivalent of R_th=10Ω, the load resistance should also be R_L=10Ω to achieve maximum power transfer.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

he maximum power transfer theorem states that the maximum power is delivered to the load when the load resistance R_L is equal to the Thevenin resistance R_th.
The formula for the maximum power delivered to the load is:

Where,
P_max is the maximum power delivered to the load,
V_th is the Thevenin voltage,
R_th is the Thevenin resistance.

Given that,
V_th=30V,
R_th=15Ω.

Now, substituting these values into the formula,

P_max = 30²/(4×15)
 ​       = 900/60
       =15
∴P_max=15W

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel