Definition of Electric Circuits:
An electric circuit is a closed loop or pathway that allows the flow of electric current. It consists of various electrical components, such as resistors, capacitors, inductors, and voltage sources, connected together through conductors.
Formulas in Electric Circuits:
Ohm's Law
V = I * R
This formula relates the voltage (V) across a resistor, the current (I) flowing through it, and the resistance (R) of the resistor.
Problem solved: Ohm's Law
Given:
Resistance (R) = 10 Ω
Current (I) = 2 A
Find: Voltage (V)
Solution:
Using Ohm's Law: V = I * R
Substituting the given values: V = 2 * 10
V = 20 V
Therefore, the voltage across the resistor is 20 Volts.
Power Law
P = V * I
This formula calculates the power (P) dissipated by a component, given the voltage (V) across it and the current (I) flowing through it.
Problem solved: Power Law
Given:
Voltage (V) = 12 V
Current (I) = 3 A
Find: Power (P)
Solution:
Using the Power Law: P = V * I
Substituting the given values: P = 12 * 3
P = 36 W
Therefore, the power dissipated by the component is 36 Watts.
Series Resistance
Rs = R1 + R2 + R3 + ...
In a series circuit, the total resistance (Rs) is equal to the sum of individual resistances (R1, R2, R3, etc.).
Problem solved: Series Resistance
Given:
Resistor 1 (R1) = 5 Ω
Resistor 2 (R2) = 10 Ω
Resistor 3 (R3) = 15 Ω
Find: Total resistance (Rs)
Solution:
Using the Series Resistance formula: Rs = R1 + R2 + R3
Substituting the given values: Rs = 5 + 10 + 15
Rs = 30 Ω
Therefore, the total resistance in the series circuit is 30 Ohms.
Parallel Resistance
Rp = 1 / (1/R1 + 1/R2 + 1/R3 + ...)
In a parallel circuit, the reciprocal of the total resistance (Rp) is equal to the sum of the reciprocals of individual resistances (R1, R2, R3, etc.).
Problem solved: Parallel Resistance
Given:
Resistor 1 (R1) = 3 Ω
Resistor 2 (R2) = 6 Ω
Find: Total resistance (Rp)
Solution:
Using the Parallel Resistance formula: Rp = 1 / (1/R1 + 1/R2)
Substituting the given values: Rp = 1 / (1/3 + 1/6)
Rp = 1 / (2/6 + 1/6)
Rp = 1 / (3/6)
Rp = 2 Ω
Therefore, the total resistance in the parallel circuit is 2 Ohms.
Kirchhoff's Laws
a. Kirchhoff's Voltage Law (KVL): The sum of voltages in any closed loop of a circuit is equal to zero.
b. Kirchhoff's Current Law (KCL): The sum of currents entering a node is equal to the sum of currents leaving the node.
Problem solved: Kirchhoff's Voltage Law (KVL)
Given:
Voltage 1 (V1) = 10 V
Voltage 2 (V2) = 5 V
Voltage 3 (V3) = 8 V
Find: Sum of voltages (V_total)
Solution:
Using Kirchhoff's Voltage Law: V_total = V1 + V2 + V3
Substituting the given values: V_total = 10 + 5 + 8
V_total = 23 V
Therefore, the sum of voltages in the closed loop of the circuit is 23 Volts.
More electric circuit problems with solutions
Problem 1:
Calculate the current flowing through a resistor with a voltage of 12 V and a resistance of 4 Ω.
Solution:
Using Ohm's Law: V = I * R
Substituting the given values: 12 = I * 4
Solving for I: I = 12 / 4 = 3 A
Therefore, the current flowing through the resistor is 3 Amps.
Problem 2:
Determine the power dissipated by a component with a voltage of 9 V and a current of 2 A.
Solution:
Using the Power Law: P = V * I
Substituting the given values: P = 9 * 2 = 18 W
Therefore, the power dissipated by the component is 18 Watts.
Problem 3:
In a series circuit, there are three resistors with values of 5 Ω, 10 Ω, and 15 Ω. Calculate the total resistance.
Solution:
Using the Series Resistance formula: Rs = R1 + R2 + R3
Substituting the given values: Rs = 5 + 10 + 15 = 30 Ω
Therefore, the total resistance in the series circuit is 30 Ohms.
Problem 4:
In a parallel circuit, there are two resistors with values of 3 Ω and 6 Ω. Calculate the total resistance.
Solution:
Using the Parallel Resistance formula: Rp = 1 / (1/R1 + 1/R2)
Substituting the given values: Rp = 1 / (1/3 + 1/6) = 2 Ω
Therefore, the total resistance in the parallel circuit is 2 Ohms.
Problem 5:
In a circuit, three resistors with values of 4 Ω, 6 Ω, and 8 Ω are connected in parallel. Calculate the equivalent resistance.
Solution:
Substituting the given values: Rp = 1 / (1/4 + 1/6 + 1/8)
To simplify the calculation, we can find a common denominator:
Rp = 1 / (3/12 + 2/12 + 3/12)
Rp = 1 / (8/12)
Rp = 12/8
Rp = 1.5 Ω
Therefore, the equivalent resistance in the parallel circuit is 1.5 Ohms.
No comments:
Post a Comment