Series Circuits & Parallel Networks - GATE EE Network Theory Free MCQ Test

In a series circuit, the current across all elements remain the same and the total volt-age of the circuit is the sum of the voltages across all the elements.

• The 60 ohm resistance is shorted since current always chooses the low resistance path. So, whole of the current will flow through the alternative route and there will be 0 current through 60Ω resistance.
• As, Voltage across short circuit is equal to zero, hence voltage across the resistor is 0.

Solution:

Given the circuit, we need to find the equivalent resistance between points A and B.

Step 1: Identify series combinations

• Resistors R₁ = 3 Ω and R₂ = 2 Ω are in series, so their combined resistance is:

R_a = R₁ + R₂ = 3 + 2 = 5 Ω

• Resistors R₄ = 6 Ω and R₅ = 4 Ω are also in series:

R_b = R₄ + R₅ = 6 + 4 = 10 Ω

Step 2: Identify parallel combinations

Now, resistors R_a (5 Ω), R₃ (30 Ω), and R_b (10 Ω) are connected in parallel between points A and B.

The formula for equivalent resistance of parallel resistors is:

1 / R_eq = 1 / R_a + 1 / R₃ + 1 / R_b

Substitute the values:

1 / R_eq = 1 / 5 + 1 / 30 + 1 / 10

Calculate the sum:

1 / R_eq = (6/30) + (1/30) + (3/30) = 10/30 = 1/3

Therefore,

R_eq = 3 Ω

Answer: The equivalent resistance between points A and B is 3 Ω.

Total current = 150/(6+12+15) = 4.55A.

V across 6 ohm = Total current x resistance

= 4.55 * 6 = 27.24V.

Since the two bulbs are connected in series, if the first bulb burns out there is a break in the circuit and hence the second bulb does not glow.

V=IR hence voltage across a series resistor circuit is proportional to the value of the resistance because current remains same in series circuit.

In a series circuit, the current remains the same across all resistors hence the voltage divides proportionally among all resistors.

A short is just a wire. The potential difference between two points of a wire is zero hence the voltage measured is equal to zero.

Current is inversely proportional to resistance so resistance increased 2 times means current is decreased 2 times.

In a parallel circuit, each bulb is connected across the same voltage source, and the current flowing through each bulb is independent of the others. If one bulb blows out, it does not affect the current flow through the other bulb. The remaining bulb will continue to glow with the same brightness because the voltage across it remains unchanged.

• In parallel circuits, the failure of one component (like a bulb) does not interrupt the operation of the other components, unlike in series circuits where the failure of one component stops the current flow through the entire circuit.

I = V/R. Since in parallel circuit, voltage is same across all resistors. Hence, across the 20 ohm resistor, V = 200V so I = 200/20  = 10A

 In parallel circuits, the current across the circuits vary whereas the voltage remains the same.

The 1 Ω and 2 Ω resistors are in series; their combined resistance is 1 Ω + 2 Ω = 3 Ω.

This 3 Ω group is in parallel with the 3 Ω resistor. The equivalent resistance of two resistors in parallel is R_eq = (R1 × R2) / (R1 + R2).

So R_eq = (3 Ω × 3 Ω) / (3 Ω + 3 Ω) = 9/6 Ω = 1.5 Ω.

These are in series with 4 Ω and 5 Ω, so total resistance is R_total = 1.5 Ω + 4 Ω + 5 Ω = 10.5 Ω.

Current I = V / R_total = 120 V / 10.5 Ω = 11.42 A. This corresponds to option B.

The voltage across all branches in a parallel circuit is the same as that of the source voltage. Hence the voltage across the 10 ohm resistor and the open circuit is the same=100V.

The voltage across a short is always equal to zero whether it is connected in series or parallel.

The total current leaving a node is the same as the current that enters it. Total I=I₁+I₂+I₃=3+4+5=12A.

The resistors are connected in parallel, hence the equivalent resistance= 1/(1/20=1/20+1/20+1/20)=5ohm.

Batteries can be connected in series or parallel depending on the desired electrical outcome. When connected in series, the voltages add up, providing a higher total voltage, which is useful when higher voltage is needed. Conversely, connecting batteries in parallel increases the total current capacity while maintaining the same voltage, which is beneficial when a higher current is required. Therefore, batteries are generally connected in either series or parallel based on the specific requirements.

 In series circuits the total resistance is the sum of all the resistance in the circuit, hence the total is greater than the largest resistance.

Since each resistor is connected across a 5V source, we calculate the current through each resistor individually.

1. Current through the 5-ohm resistor:
   $$I_1 = \frac{V}{R_1} = \frac{5 \, \text{V}}{5 \, \Omega} = 1 \, \text{A}$$

2. Current through the 3-ohm resistor:
   

Final Answer:

• The current through the 5-ohm resistor is 1 A.
• The current through the 3-ohm resistor is approximately 1.67 A.

The total current drawn from the 5V source would be the sum of these currents:

So, the correct answer for the current through the 3-ohm resistor is 1.67 A, and not 3.125 A.

• 3 and 2 ohm resistors are in series, So 3+2 = 5ohm
• 5 and 5 ohm resistors are in parallel. So 5 || 5 = 2.5ohms
• 2.5 and1.5ohm resistors are in series. So 2.5+1.5 = 4ohms
• Finally, 4 and 4ohm resistors are in parallel. So 4 || 4 = 2ohms.
  That is resistance across A and B terminals is 2 ohms

Ideal Voltage Source: An ideal voltage source has zero internal resistance.

Practical Voltage Source: A practical voltage source consists of an ideal voltage source (V_S) in series with internal resistance (R_S).

An ideal voltage source and a practical voltage source can be represented as shown in the figure.

Ideal Current Source: An ideal voltage source has infinite resistance. Infinite resistance is equivalent to zero conductance. So, an ideal current source has zero conductance.

Practical Current Source: A practical current source is equivalent to an ideal current source in parallel with a high resistance or low conductance.

Ideal and practical current sources are represented as shown the below figure.

Step 1: Identify Parallel Resistors

The circuit shows three 20Ω resistors connected in parallel.

Step 2: Calculate Equivalent Resistance of Parallel Resistors

The formula for equivalent resistance (Req) of resistors in parallel is: 1/Req = 1/R1 + 1/R2 + 1/R3

In this case, R1 = R2 = R3 = 20Ω. Therefore:

1/Req = 1/20Ω + 1/20Ω + 1/20Ω = 3/20Ω

Req = 20Ω/3 ≈ 6.67Ω

Step 3: Final Resistance

The equivalent resistance between points A and B is approximately 6.67Ω.

Final Answer

6.67 Ω

Resistances in parallel combination: When two or more resistances are connected across different branches such that the potential drop across them are the same, they are said to be in parallel connection.

The net resistance/equivalent resistance (R) of resistances in parallel is given by:
Equivalent resistance (R): 
EXPLANATION:
If n resistance connected in parallel the,

Hence,

Concept : 

Ideal current source: an ideal current source has infinite resistance. Infinite Resistance is equivalent to zero conductance. 

Practical current source: Practical current source is equivalent to the ideal current source in parallel with high resistance or low conductance.

Ideal and practical current sources are represented as given below :

Ideal current source: An ideal current source has infinite resistance. Infinite resistance is equivalent to zero conductance. So, an ideal current source has zero conductance.

Practical current source: A practical current source is equivalent to an ideal current source in parallel with high resistance or low conductance.

Ideal and practical current sources are represented as shown in the below figure:

Important Points

Ideal voltage source: An ideal voltage source have zero internal resistance.

Practical voltage source: A practical voltage source consists of an ideal voltage source (VS) in series with internal resistance (RS) as follows.

An ideal voltage source and a practical voltage source can be represented as shown in the figure.

Assuming i1 and i2 be the currents in loop 1 and 2 respectively. In loop 1, 4+2i1+3(i1-17/3)+4(i1-i2)+5=0
In loop 2, i2(4+1+5)-4i1-5=0 =>-4i1+10i2=5.
Solving these equations simultaneously i2=1.041A and i1=1.352A
V=i2*5= 5.21V.