Important Questions of Network Analysis

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Important Questions

Ques. Find the equivalent resistance for a circuit where R₁ = 15Ω, R₂ = 20Ω, and R₃ = 30Ω are in a parallel arrangement. (3 marks) 

Ans. For a parallel arrangement of resistors, the equivalent resistance is: 

1 / R_P = 1 / R₁ + 1 / R₂ + 1 / R₃

1 / R_P = 1 / 15Ω + 1 / 20Ω + 1 / 30Ω

1 / R_P = (4 + 3 + 2) / 60Ω

1 / R_P = 9 / 60Ω

1 / R_P = 3 / 20Ω

Ques. What is the equivalent resistance if 3Ω, 20Ω and 32Ω are connected in series. (2 marks) 

Ans. Equivalent resistance in series is given by

R = R₁ + R₂ + …… + R_n

= 3 Ω + 20 Ω + 32 Ω

Ques. For the given circuit below, calculate the equivalent resistance between the end points A and B. (3 marks) 

Ans.  The exp-ression for the equivalent resistance of the resistor connected in series is-

R_S = R₁ + R₂ + R₃

R_S = 2 + 3 + 4

R_S = 3

Ques. What is the equivalent resistance if 3Ω, 20Ω and 32Ω are connected in series. (2 marks) 

Ans. Equivalent resistance in series is given by

R = R₁ + R₂ + …… + R_n

= 3 Ω + 20 Ω + 32 Ω

Ques. What is the equivalent resistance if 34 Ω and 20 Ω are connected in parallel. (3 marks) 

Ans. Equivalent resistance in parallel is given by

\(\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}+.......+\frac{1}{R_n}\)

\(\frac{1}{R_p} = \frac{1}{34} + \frac{1}{20}\)

= 0.029 + 0.05

= 0.079

R_p= 12.66Ω

Ques. An EMF source of 8.0 V is connected to a resistive electrical appliance (a light bulb). An electric current of 2.0 A flows through it. Consider the conducting wires to be resistance-free. Calculate the resistance. (2 marks) 

Ans. When we are asked to determine the value of resistance when the values of voltage and current are given, we cover R in the triangle. This leaves us with V ÷ I.

R = V ÷ I

R = 8 V ÷ 2 A = 4 Ω

R = 4Ω

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Ques. What are two similarities and two differences between a circuit with 2 bulbs in series and a circuit with 2 bulbs in parallel? (3 marks) 

Ans. The differences between the circuit when joined in series and parallel are-

• The series-connected bulbs have the same current path Whereas, the Parallel bulbs have two different paths for the current.
• The series-connected bulbs have distributed source voltage but the Parallel connected bulbs have identical source voltage.

The Similarities between the circuit in series and parallel, are-

• The series and parallel connected bulbs are affected, if the source voltage varies.
• Series-connected bulbs and parallel-connected bulbs dissipate electrical energy in a circuit.

Ques. What do you Understand by the Terms Bilateral and Unilateral Network? (2 marks) 

Ans. The bilateral circuit allows the circuit to flow in both directions. The circuit properties remain the same even after changing the direction of the voltage or current.

Examples- Resistance, Capacitance, Inductance

The unilateral network is defined as the flow of current but in different directions.

Examples- Diode, vacuums etc

Ques. State ohm's law? (3 marks) 

Ans. The most basic and important law of electric circuits is the Ohm's Law. Ohm's Law links voltage (V) and current (I) to the properties of the conductor, that is its resistance (r) in a circuit. 

• Ohm’s law states that voltage across a conductor is proportional to the current flowing through it, considering all physical conditions and temperatures remain constant. 
• Ohm's Law formula is V = IR where V = voltage, I = current, and R = resistance.

Ohm's Law

Mathematically, Ohm's Law is written as:

V = I * R

Where, 

• V is the Voltage (Potential Difference) measured in Volts. 

To find voltage (V),

[V = I x R] V(volts) = (amps) x R(Ω)

• It is the current flowing through the conductor.

To Find the current (I),

[I = V ÷ R] I(amps) = V(volts) + R(Ω)

• R is the resistance of the circuit measured in Ohms (Ω).

To find the resistance (R), 

[R = V ÷ I] R= V volts) ÷ I(amps)

Ques. Explain Kirchhoff's first law? (3 marks) 

Ans. Kirchoffs current law states that the total current flowing a junction or a node is equal to charge of the node as there is no charge lost.

Kirchhoff's First Law

• According to the above diagram the currents the currents entering the node is equal to the current leaving the node
• The current I₁, I₂ I₃ enter the node as positive and I₄ I₅ exit as negative. The equation is I₁ + I₂ + I₃ (- I₄ - I₅) = 0.
• A node is a junction connecting two or more routes like cables and other components.
• Kirchhoff’s First law can be applied to analyze the parallel circuits.

Ques. If R₁ = 2Ω, R₂ = 4Ω, R₃ = 6Ω, determine the electric current that flows in the circuit below. (5 marks) 

Ans. According to the law the direction of current is always from positive terminal to negative terminal and the thing we have to keep in mind are:

• One direction needs to be chosen, here we’ll choose a clockwise direction.
• Current will flow across the resistor. Hence, V = IR is negative.
• When the current moves from low to high, then the electromotive source is signed positive because of the energy charging at the source. 
• When the current moves from high to low voltage (+ to -), then the source is negative because of the emptying of energy at the emf source.

The Direction of the current is the same as the direction of clockwise rotation.

 – IR₁ + E₁ – IR₂ – IR₃ – E₂ = 0

After Substituting the values in the equation,

–2I + 10 – 4I – 6I – 5 = 0

-12I + 5 = 0

I = -5/-12

I = 0.416 A

• The electric current that flows in the circuit is 0.416 A. 
• The electric current is positive which shows that the direction of the electric current is the same as the direction of clockwise rotation. 

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