Mastering Circuits!
Mastering the Basic Laws of Electrical Circuit is essential for everyone. Today, we are going to learn about the Basic Laws like Ohm’s Law, Nodes, Branches and Loops, Kirchhoff’s Laws (KVL, KCL), Series Resistors & Voltage Division, Parallel Resistors & Current Division, Wye-Delta Transformations. We will also solve some problems related to these topics. Learn these topics with Mastering Circuits.
Ohm’s Law
Ohm’s Law: Ohm’s law states that the voltage v across a resistor is directly proportional to the current i flowing through the resistor.
That is, \small \begin{aligned} v &\propto i \\ \Rightarrow v &= iR \\ \therefore R &= \frac{v}{i} \end{aligned}
A short circuit is a circuit element with resistance approaching zero.
v = iR = 0
Figure 1 – Short Circuit (R = 0)
An open circuit is a circuit element with resistance approaching infinity.
\small i = \lim_{R \rightarrow \infty} \frac{v}{R} = 0
Figure 2 – Open Circuit (R = \infty)
A useful quantity in circuit analysis is the reciprocal of resistance R , known as conductance and denoted by G :
\small G = \frac{1}{R} = \frac{i}{v}Conductance is the ability of an element to conduct electric current. It is measured in mhos (\mho) or siemens (S) .
As, \small \begin{aligned} &G = \frac{i}{v} \therefore i = Gv \end{aligned}
The power dissipated by a resistor can be expressed in terms of R as:
We know, \small \begin{aligned} &p = vi \\ &\Rightarrow p = \frac{v^2}{R} \quad \left[ \because i = \frac{v}{R} \right] \\ &\Rightarrow p = i^2R \quad \left[ \because v = iR \right] \\ &\therefore p = vi = \frac{v^2}{R} = i^2R \end{aligned}
The power dissipated by a resistor may also be expressed in terms of G as:
\small \begin{aligned} \therefore p = vi = v^2G = \frac{i^2}{G} \end{aligned}Solved Problems on Ohm’s Law
Problem 1
Pb-1: An electric iron draws 2 \,A at 120 \,V . Find its resistance.
Solution:
From Ohm’s Law, \small \begin{aligned} R &= \frac{v}{i} \\ \Rightarrow R &= \frac{120}{2} \\ \therefore R &= 60 \,\Omega \end{aligned}
Problem 2
Pb-2: The essential component of a toaster is an electrical element (a resistor) that converts electrical energy to heat energy. How much current is drawn by a toaster with resistance 15 \,\Omega at 110 \,V ?
Solution:
From Ohm’s Law, \small \begin{aligned} i &= \frac{v}{R} \\ \Rightarrow i &= \frac{110}{15} \\ \therefore i &= 7.33 \,A \end{aligned}
Problem 3
Pb-3: In the circuit shown in Fig. 3, calculate the current i , the conductance G , and the power p .
Figure 3 – Circuit diagram for Pb-3
Solution:
From the Fig. 3, \small \begin{aligned} i &= \frac{v}{R} = \frac{30}{5 \times 10^3} = 6 \times 10^{-3} \,A = 6 \,mA \\ G &= \frac{1}{R} = \frac{1}{5 \times 10^3} = 2 \times 10^{-4} \,S = 0.2 \,mS \\ p &= vi = 30 \times 6 \times 10^{-3} = 0.18 \,W = 180 \,mW \end{aligned}
Problem 4
Pb-4: For the circuit shown in Fig. 4, calculate the voltage v , the conductance G , and the power p .
Figure 4 – Circuit diagram for Pb-4
Solution:
From the Fig. 4,
\small \begin{aligned} v &= iR = 3 \times 10^{-3} \times 10 \times 10^3 = 30 \,V \\ G &= \frac{1}{R} = \frac{1}{10^4} = 10^{-4} \,S = 100 \,\mu S \\ p &= vi = 30 \times 3 \times 10^{-3} = 0.09 \,W = 90 \,mW \end{aligned}Problem 5
Pb-5: A voltage source of 20 \sin \pi t \,V is connected across a 5 \,k\Omega resistor. Find the current through the resistor and the power dissipated.
Solution:
\small \begin{aligned} i &= \frac{v}{R} = \frac{20 \sin \pi t}{5 \times 10^3} = 4 \times 10^{-3} \sin \pi t \,A \\ &= 4 \sin \pi t \ mA \\ p &= vi = 20 \sin \pi t \times 4 \times 10^{-3} \sin \pi t \\ &= 80 \times 10^{-3} \sin^2 \pi t \,W \\ &= 80 \sin^2 \pi t \ mW \end{aligned}Problem 6
Pb-6: A resistor absorbs an instantaneous power of 30 \cos^2 t \ mW when connected to a voltage source 15 \cos t \,V . Find i and R .
Solution:
\small \begin{aligned} p &= vi \Rightarrow 30 \cos^2 t = 15 \cos t \times i \\ \Rightarrow i &= \frac{30 \cos^2 t}{15 \cos t} \\ \therefore i &= 2 \cos t \ mA \\ R &= \frac{v}{i} = \frac{15 \cos t}{2 \cos t \times 10^{-3}} = 7500 \,\Omega = 7.5 \ k\Omega \end{aligned}Nodes, Branches, and Loops
Understanding Nodes, Branches, and Loops is necessary because they form the fundamental framework for applying Kirchhoff’s Laws (KCL, KVL).
Figure 5 – Nodes, Branches, and Loops
Branch: A branch represents a single element such as a voltage source or a resistor.
Node: A node is the point of connection between two or more branches.
Loop: A loop is any closed path in a circuit.
A network with b branches, n nodes, and l independent loops will satisfy the fundamental theorem of network topology:
b = l+n-1where, \begin{aligned} b &= \text{branches} \\ l &= \text{loops} \\ n &= \text{nodes} \end{aligned}
According to network topology:
Two or more elements are in series if they exclusively share a single node and consequently carry the same current.
Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them.
Solved Problems on Nodes, Branches, and Loops
Problem 7
Pb-7: Determine the number of branches and nodes in the circuit shown in the Fig. 6. Identify which elements are in series and which are in parallel.
Figure 6 – Circuit diagram for Pb-7
Solution:
Figure 7 – Circuit diagram for solving the Pb-7
Since there are 4 elements in the circuit, the circuit has 4 branches: 10 \ V , 5 \ \Omega , 6 \ \Omega , and 2 \ A .
The circuit has 3 nodes as identified in Fig. 7.
The 5 \ \Omega resistor is in series with the 10 \ V voltage source because the same current would flow in both.
The 6 \ \Omega resistor is in parallel with the 2 \ A current source because both are connected to the same nodes 2 and 3 .
Problem 8
Pb-8: How many branches and nodes does the circuit in Fig. 8 have? Identify the elements that are in series and in parallel.
Figure 8 – Circuit diagram for Pb-8
Solution:
Figure 9 – Circuit diagram for solving the Pb-8
Since there are 5 elements in the circuit, the circuit has 5 branches: 10 \ V , 1 \ \Omega , 2 \ \Omega , 4 \ \Omega , and 5 \ \Omega .
The circuit has 3 nodes as identified in Fig. 9.
Combination 1: The 1 \ \Omega and 2 \ \Omega resistors are in parallel.
Combination 2: The 4 \ \Omega resistor and 10 \ V voltage source are also in parallel.
Combination 1 & 2 are in series with the 5 \ \Omega resistor.
Kirchhoff’s Laws
KCL: Kirchhoff’s Current Law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero.
Mathematically, \small \begin{aligned} \sum_{n=1}^{N} i_n = 0 \end{aligned}
Figure 10 – Currents at a node illustrating KCL
\begin{aligned} &i_1 + (-i_2) + i_3 + i_4 + (-i_5) = 0 \\ &\therefore i_1 + i_3 + i_4 = i_2 + i_5 \end{aligned}The sum of currents entering a node is equal to the sum of currents leaving the node.
KVL: Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero.
Mathematically, \small \begin{aligned} \sum_{m=1}^{M} v_m = 0 \end{aligned}
Figure 11 – A single-loop circuit illustrating KVL
\begin{aligned} &-v_1 + v_2 + v_3 - v_4 + v_5 = 0 \\ &\therefore v_2 + v_3 + v_5 = v_1 + v_4 \end{aligned} \small \text{Sum of voltage drops} = \text{Sum of voltage rises}Solved Problems on Kirchhoff’s Laws
Problem 9
Pb-9: For the circuit in Fig. 12, find voltages v_1 and v_2 .
Figure 12 – Circuit diagram for Pb-9
Solution:
Figure 13 – Circuit diagram for solving the Pb-9
\small \begin{aligned} &\text{From Ohm's Law,} \\ &\qquad \qquad \qquad v_1 = 2i \ ; \ v_2 = -3i \\ &\text{Applying KVL around the loop,} \\ &\qquad \qquad \qquad v_1 - v_2 - 20 = 0 \Rightarrow 2i + 3i = 20 \\ &\qquad \qquad \qquad \therefore i_1 = 4 \ A \\ &\qquad \qquad \quad \therefore v_1 = 8 \ V \ , \ v_2 = -12 \ V \end{aligned}Problem 10
Pb-10: Find v_1 and v_2 in the circuit of Fig. 14.
Figure 14 – Circuit diagram for Pb-10
Solution:
Figure 15 – Circuit diagram for solving the Pb-10
\small \begin{aligned} &\text{From Ohm's Law,} \\ &\qquad \qquad \qquad v_1 = 4i \ ; \ v_2 = -2i \\ &\text{Applying KVL around the loop,} \\ &\qquad \qquad \qquad -32 + v_1 + 8 - v_2 = 0 \\ &\qquad \qquad \qquad \Rightarrow 4i + 2i = 24 \\ &\qquad \qquad \qquad \therefore i = 4 \ A \\ &\qquad \qquad \quad \therefore v_1 = 16 \ V \ ; \ v_2 = -8 \ V \end{aligned}Problem 11
Pb-11: Determine v_0 and i in the circuit shown in Fig. 16.
Figure 16 – Circuit diagram for Pb-11
Solution:
Figure 17 – Circuit diagram for solving the Pb-11
\small \begin{aligned} &\text{Applying KVL around the loop,} \\ &\qquad \qquad \qquad -12 + 4i + 2v_0 - 4 + 6i = 0 \\ &\qquad \qquad \qquad \Rightarrow 4i + 2 \cdot (-6i) + 6i = 16 \\ &\qquad \qquad \qquad \Rightarrow 4i - 12i + 6i = 16 \Rightarrow -2i = 16 \\ &\qquad \qquad \qquad \therefore i = -8 \ A \\ &\qquad \qquad \quad \therefore v_0 = -6i = 48 \ V \end{aligned}Problem 12
Pb-12: Find v_x and v_0 in the circuit of Fig. 18.
Figure 18 – Circuit diagram for Pb-12
Solution:
Figure 19 – Circuit diagram for solving the Pb-12
\small \begin{aligned} &\text{Applying KVL around the loop,} \\ &\qquad \qquad \qquad -70 + 10i + 2v_x + 5i = 0 \\ &\qquad \qquad \qquad \Rightarrow 10i + 2 \cdot (10i) + 5i = 70 \\ &\qquad \qquad \qquad \Rightarrow 35i = 70 \ \therefore i = 2 \ A \\ &\qquad \qquad \quad \therefore v_x = 10i = 20 \ V \\ &\qquad \qquad \quad \& \ v_0 = -5i = -10 \ V \end{aligned}Problem 13
Pb-13: Find current i_0 and voltage v_0 in the circuit shown in Fig. 20.
Figure 20 – Circuit diagram for Pb-13
Solution:
Figure 21 – Circuit diagram for solving the Pb-13
\small \begin{aligned} &\text{KCL at node a,} \\ &\qquad \qquad \qquad 0.5i_0 - i_0 + 3 = 0 \\ &\qquad \qquad \qquad \Rightarrow -0.5i_0 = -3 \\ &\qquad \qquad \qquad \therefore i_0 = 6 \ A \\ &\qquad \qquad \quad \therefore v_0 = 4i_0 = 24 \ V \end{aligned}Problem 14
Pb-14: Find v_0 and i_0 in the circuit of Fig. 22.
Figure 22 – Circuit diagram for Pb-14
Solution:
Figure 23 – Circuit diagram for solving the Pb-14
\small \begin{aligned} &\text{KCL at node a,} \\ &\qquad \qquad \qquad 9 - i_0 - \frac{i_0}{4} - \frac{v_0}{8} = 0 \\ &\qquad \qquad \qquad \Rightarrow 9 - i_0 - \frac{i_0}{4} - \frac{2i_0}{8} = 0 \\ &\qquad \qquad \qquad \Rightarrow \frac{-8i_0 - 2i_0 - 2i_0}{8} = -9 \\ &\qquad \qquad \qquad \Rightarrow -12i_0 = -72 \ \therefore i_0 = 6 \ A \\ &\qquad \qquad \quad \therefore v_0 = 2i_0 = 12 \ V \end{aligned}Problem 15
Pb-15: Find currents and voltages in the circuit shown in Fig. 24.
Figure 24 – Circuit diagram for Pb-15
Solution:
Figure 25 – Circuit diagram for solving the Pb-15
\small \begin{aligned} &\text{KCL at node a,} \\ &\qquad \qquad \qquad i_1 - i_2 - i_3 = 0 \ ---(i) \\ &\text{KVL around loop 1,} \\ &\qquad \qquad \qquad -30 + 8i_1 + 3i_2 = 0 \ \\ &\qquad \qquad \qquad \therefore 8i_1 + 3i_2 = 30 \ ---(ii) \\ &\text{KVL around loop 2,} \\ &\qquad \qquad \qquad 6i_3 - 3i_2 = 0 \ \\ &\qquad \qquad \qquad \therefore -3i_2 + 6i_3 = 0 \ ---(iii) \\ &\text{Solving (i), (ii) \& (iii),} \\ &\qquad \qquad \qquad \therefore i_1 = 3 \ A \ , \ i_2 = 2 \ A \ , \ i_3 = 1 \ A \\ &\qquad \qquad \quad \therefore v_1 = 24 \ V \ , \ v_2 = 6 \ V \ , \ v_3 = 6 \ V \end{aligned}Problem 16
Pb-16: Find the currents and voltages in the circuit shown in Fig. 26.
Figure 26 – Circuit diagram for Pb-16
Solution:
Figure 27 – Circuit diagram for solving the Pb-16
\small \begin{aligned} &\text{KCL at node a,} \\ &\qquad \qquad \qquad i_1 - i_2 - i_3 = 0 \ ---(i) \\ &\text{KVL around loop 1,} \\ &\qquad \qquad \qquad -10 + 2i_1 + 8i_2 = 0 \ \\ &\qquad \qquad \qquad \therefore 2i_1 + 8i_2 = 10 \ ---(ii) \\ &\text{KVL around loop 2,} \\ &\qquad \qquad \qquad 4i_3 - 6 - 8i_2 = 0 \ \\ &\qquad \qquad \qquad \therefore -8i_2 + 4i_3 = 6 \ ---(iii) \\ &\text{Solving (i), (ii) \& (iii),} \\ &\qquad \qquad \qquad \therefore i_1 = 3 \ A \ , \ i_2 = 0.5 \ A \ , \ i_3 = 2.5 \ A \\ &\qquad \qquad \quad \therefore v_1 = 6 \ V \ , \ v_2 = 4 \ V \ , \ v_3 = 10 \ V \end{aligned}Series Resistors & Voltage Division
The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances.
\small \begin{aligned} R_{eq} = R_1 + R_2 + ------ R_N \end{aligned}
Figure 28 – A single-loop circuit with two resistors in series
Voltage Division: The voltage division rule states that the voltage across any of the series components in a series circuit is equal to the product of value of that resistance and the total supply voltage, divided by the total resistance of the series circuit.
Figure 29 – Resistors R_1 , R_2 , and R_3 connected in series showing individual voltage drops v_1 , v_2 , and v_3
\small \begin{aligned} v_1 &= iR_1 = \frac{v}{R_{eq}} \cdot R_1 & i &= \frac{v}{R_{eq}} \\ \therefore v_1 &= \frac{R_1}{R_{eq}} \cdot v & \therefore i &= \frac{v}{R_1+R_2+R_3} \\ v_2 &= iR_2 = \frac{v}{R_{eq}} \cdot R_2 \\ \therefore v_2 &= \frac{R_2}{R_{eq}} \cdot v \\ v_3 &= iR_3 = \frac{v}{R_{eq}} \cdot R_3 \\ \therefore v_3 &= \frac{R_3}{R_{eq}} \cdot v \end{aligned} \small \begin{aligned} &\text{Therefore,} \\ &\qquad \qquad \qquad v_1 = \frac{R_1}{R_{eq}} \cdot v \\ &\qquad \qquad \qquad v_2 = \frac{R_2}{R_{eq}} \cdot v \\ &\qquad \qquad \qquad v_3 = \frac{R_3}{R_{eq}} \cdot v \end{aligned} \small \begin{aligned} &\text{In general,} \\ &\qquad \qquad \qquad v_x = \frac{R_x}{R_{eq}} \cdot v_{total} \end{aligned}Parallel Resistors & Current Division
The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum.
\small \begin{aligned} \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{R_1+R_2}{R_1 \cdot R_2} \therefore R_{eq} = \frac{R_1 \cdot R_2}{R_1+R_2} \end{aligned}
Figure 30 – Two resistors in parallel
Current Division: The current division rule states that the current through any of the parallel components in a parallel circuit is equal to the product of the total resistance of the parallel circuit and the total current, divided by the value of that resistance.
Figure 31 – Resistors R_1 , R_2 , and R_3 connected in parallel showing individual current flows i_1 , i_2 , and i_3
\small \begin{aligned} v &= i_1R_1 \Rightarrow i_1 = \frac{v}{R_1} = \frac{iR_{eq}}{R_1} & i &= \frac{v}{R_{eq}} \\ \therefore i_1 &= \frac{R_{eq}}{R_1} \cdot i & \therefore v &= iR_{eq} \\ v &= i_2R_2 \Rightarrow i_2 = \frac{v}{R_2} = \frac{iR_{eq}}{R_2} \\ \therefore i_2 &= \frac{R_{eq}}{R_2} \cdot i \\ v &= i_3R_3 \Rightarrow i_3 = \frac{v}{R_3} = \frac{iR_{eq}}{R_3} \\ \therefore i_3 &= \frac{R_{eq}}{R_3} \cdot i \end{aligned} \small \begin{aligned} &\text{Therefore,} \\ &\qquad \qquad \qquad i_1 = \frac{R_{eq}}{R_1} \cdot i \\ &\qquad \qquad \qquad i_2 = \frac{R_{eq}}{R_2} \cdot i \\ &\qquad \qquad \qquad i_3 = \frac{R_{eq}}{R_3} \cdot i \end{aligned} \small \begin{aligned} &\text{In general,} \\ &\qquad \qquad \qquad i_x = \frac{R_{eq}}{R_x} \cdot i_{total} \end{aligned}The equivalent conductance G_{eq} of N resistors in series:
\small \begin{aligned} \frac{1}{G_{eq}} = \frac{1}{G_1} + \frac{1}{G_2} + \frac{1}{G_3} + --- + \frac{1}{G_N} \end{aligned}The equivalent conductance G_{eq} of N resistors in parallel:
\small \begin{aligned} G_{eq} = G_1 + G_2 + G_3 + --- + G_N \end{aligned}Solved Problems on Series & Parallel Resistors
Problem 17
Pb-17: Find R_{eq} for the circuit in the Fig. 32.
Figure 32 – Circuit diagram for Pb-17
Solution:
\small \begin{aligned} &R_{eq} = (((3||6) + 2)||(1 + 5)) + 4 + 8 \\ &\Rightarrow R_{eq} = \left( \Big( (3^{-1} + 6^{-1})^{-1} + 2 \Big)^{-1} \right. \\ &\qquad \qquad \left. \vphantom{\Big(} + (1 + 5)^{-1} \right)^{-1} + 4 + 8 \\ &\therefore R_{eq} = 14.4 \ \Omega \end{aligned}Problem 18
Pb-18: Find R_{eq} for the circuit in the Fig. 33.
Figure 33 – Circuit diagram for Pb-18
Solution:
\small \begin{aligned} &R_{eq} = ((((4 + 5 + 3)||4) + 3)||6) + 4 + 3 \\ &\Rightarrow R_{eq} = \left( \Big( \big( (4 + 5 + 3)^{-1} + 4^{-1} \big)^{-1} + 3 \Big)^{-1} \right. \\ &\qquad \qquad \left. \vphantom{\Big(} + 6^{-1} \right)^{-1} + 4 + 3 \\ &\therefore R_{eq} = 10 \ \Omega \end{aligned}Problem 19
Pb-19: Calculate the equivalent resistance R_{ab} in the circuit of Fig. 34.
Figure 34 – Circuit diagram for Pb-19
Solution:
\small \begin{aligned} &R_{ab} = ((((1 + 5)||4||12) + 1)||3||6) + 10 \\ &\Rightarrow R_{ab} = \left( \Big( \big( (1 + 5)^{-1} + 4^{-1} + 12^{-1} \big)^{-1} + 1 \Big)^{-1} \right. \\ &\qquad \qquad \left. \vphantom{\Big(} + 3^{-1} + 6^{-1} \right)^{-1} + 10 \\ &\therefore R_{ab} = 11.2 \ \Omega \end{aligned}Problem 20
Pb-20: Find R_{ab} for the circuit in Fig. 35.
Figure 35 – Circuit diagram for Pb-20
Solution:
\small \begin{aligned} &R_{ab} = (((((20||5) + 1)||20) + 2)||9||18) + 16 \\ &\Rightarrow R_{ab} = \Bigg( \bigg( \Big( \big( (20^{-1} + 5^{-1})^{-1} + 1 \big)^{-1} + 20^{-1} \Big)^{-1} \\ &\qquad \qquad + 2 \bigg)^{-1} + 9^{-1} + 18^{-1} \Bigg)^{-1} + 16 \\ &\therefore R_{ab} = 19 \ \Omega \end{aligned}Problem 21
Pb-21: Calculate G_{eq} in the circuit of Fig. 36.
Figure 36 – Circuit diagram for Pb-21
Solution:
\small \begin{aligned} &G_{eq} = ((8 + 12)||5) + 6 \\ &\Rightarrow G_{eq} = \left( (8 + 12)^{-1} + 5^{-1} \right)^{-1} + 6 \\ &\therefore G_{eq} = 10 \ S \end{aligned}Problem 22
Pb-22: Calculate G_{eq} for the circuit in Fig. 37.
Figure 37 – Circuit diagram for Pb-22
Solution:
\small \begin{aligned} &G_{eq} = ((8 + 4)||(2 + 4)) \\ &\Rightarrow G_{eq} = \left( (8 + 4)^{-1} + (2 + 4)^{-1} \right)^{-1} \\ &\therefore G_{eq} = 4 \ S \end{aligned}Problem 23
Pb-23: Find i_0 and v_0 in the circuit shown in Fig. 38. Calculate power dissipated in the 3 \ \Omega resistor.
Figure 38 – Circuit diagram for Pb-23
Solution:
\small \begin{aligned} &R_{eq} = (6||3) + 4 = (6^{-1} + 3^{-1})^{-1} + 4 = 6 \ \Omega \\ &\therefore i = \frac{12}{R_{eq}} = \frac{12}{6} = 2 \ A \\ &\text{According to current division rule,} \\ &\qquad i_0 = \frac{(6||3)}{3} \times i = \frac{(6^{-1} + 3^{-1})^{-1}}{3} \times 2 \\ &\qquad \therefore i_0 = \frac{4}{3} \ A \\ &\therefore v_0 = i_0 \times 3 = \frac{4}{3} \times 3 = 4 \ V \\ &\therefore p_{3\Omega} = v_0 \times i_0 = 4 \times \frac{4}{3} = \frac{16}{3} = 5.33 \ W \end{aligned}Problem 24
Pb-24: Find v_1 and v_2 in the circuit shown in Fig. 39. Also calculate i_1 and i_2 and the power dissipated in the 12 \ \Omega and 40 \ \Omega resistor.
Figure 39 – Circuit diagram for Pb-24
Solution:
\small \begin{aligned} &R_{eq} = (6||12) + (10||40) \\ &\Rightarrow R_{eq}= (6^{-1} + 12^{-1})^{-1} + (10^{-1} + 40^{-1})^{-1} \\ &\therefore R_{eq} = 12 \ \Omega \\ &\therefore i = \frac{30}{R_{eq}} = \frac{30}{12} = 2.5 \ A \\ &\text{According to current division rule,} \\ &\qquad i_1 = \frac{(6||12)}{12} \times i = \frac{(6^{-1} + 12^{-1})^{-1}}{12} \times 2.5 \\ &\qquad \therefore i_1 = 0.833 \ A \\ &\qquad i_2 = \frac{(10||40)}{40} \times i = \frac{(10^{-1} + 40^{-1})^{-1}}{40} \times 2.5 \\ &\qquad \therefore i_2 = 0.5 \ A \\ &\therefore v_1 = i_1 \times 12 = 0.833 \times 12 = 9.996 \ V \\ &\therefore v_2 = i_2 \times 40 = 0.5 \times 40 = 20 \ V \\ &\therefore p_{12\Omega} = v_1 \times i_1 = 9.996 \times 0.833 = 8.327 \ W \\ &\therefore p_{40\Omega} = v_2 \times i_2 = 20 \times 0.5 = 10 \ W \end{aligned}Problem 25
Pb-25: For the circuit shown in Fig. 40, determine: (a) the voltage v_0 , (b) the power supplied by the current source, (c) the power absorbed by each resistor.
Figure 40 – Circuit diagram for Pb-25
Solution:
Simplifying the circuit,
Figure 41 – Circuit diagram for solving the Pb-25
\small \begin{aligned} &R_{eq} = (9||18) = (9^{-1} + 18^{-1})^{-1} = 6 \ k\Omega = 6000 \ \Omega \\ &\text{According to current division rule,} \\ &\qquad i_1 = \frac{R_{eq}}{9000} \times (30 \times 10^{-3}) \\ &\qquad \Rightarrow i_{1} = \frac{6000}{9000} \times (30 \times 10^{-3}) \\ &\qquad \therefore i_1 = 0.02 \ A \\ &\qquad i_2 = \frac{R_{eq}}{18000} \times (30 \times 10^{-3}) \\ &\qquad \Rightarrow i_{2} = \frac{6000}{18000} \times (30 \times 10^{-3}) \\ &\qquad \therefore i_2 = 0.01 \ A \\ &v_0 = i_1 \times 9000 = 0.02 \times 9000 \\ &\therefore v_0 = 180 \ V \end{aligned} \small \begin{aligned} &p = v_0 \times (30 \times 10^{-3}) = 180 \times (30 \times 10^{-3}) \\ &\therefore p = 5.4 \ W \\ &\therefore p_{9k\Omega} = v_0 \times i_1 = 180 \times 0.02 = 3.6 \ W \\ &\text{As,} \\ &\qquad p = \frac{v^2}{R_{eq}} \\ &\qquad \Rightarrow v^2 = p \times R_{eq} \\ &\qquad \Rightarrow v = \sqrt{p \times R_{eq}} = \sqrt{5.4 \times 6000} \\ &\qquad \Rightarrow v = 180 \ V \ \therefore v_{total} = 180 \ V \\ &\text{Applying voltage division rule in Fig. 40,} \\ &\qquad v_{6k\Omega} = \frac{6}{6+12} \times v_{total} = \frac{6}{18} \times 180 = 60 \ V \\ &\qquad v_{12k\Omega} = \frac{12}{6+12} \times v_{total} = \frac{12}{18} \times 180 = 120 \ V \\ &\therefore p_{6k\Omega} = \frac{(v_{6k\Omega})^2}{6000} = \frac{(60)^2}{6000} = 0.6 \ W \\ &\therefore p_{12k\Omega} = \frac{(v_{12k\Omega})^2}{12000} = \frac{(120)^2}{12000} = 1.2 \ W \end{aligned} \small \begin{aligned} &\text{Therefore,} \\ &\quad \text{the voltage: } v_0 = 180 \ V \\ &\quad \text{the power supplied by the current source} \\ &\quad \text{is: } p = 5.4 \ W \\ &\quad \text{the power absorbed by } 9 \ k\Omega \text{ is: } p_{9k\Omega} = 3.6 \ W \\ &\quad \text{the power absorbed by } 6 \ k\Omega \text{ is: } p_{6k\Omega} = 0.6 \ W \\ &\quad \text{the power absorbed by } 12 \ k\Omega \text{ is: } p_{12k\Omega} = 1.2 \ W \end{aligned}Problem 26
Pb-26: For the circuit shown in Fig. 42, find: (a) v_1 and v_2 , (b) the power dissipated in the 3 \ k\Omega and 20 \ k\Omega resistors, and (c) the power supplied by the current source.
Figure 42 – Circuit diagram for Pb-26
Solution:
Simplifying the circuit,
Figure 43 – Circuit diagram for solving the Pb-26
\small \begin{aligned} &R_{eq} = ((5||20)||4) = ((5^{-1} + 20^{-1})^{-1} + 4^{-1})^{-1} \\ &\therefore R_{eq} = 2000 \ \Omega \\ &p = (30 \times 10^{-3})^2 \times R_{eq} = (30 \times 10^{-3})^2 \times 2000 \\ &\therefore p = 1.8 \ W \\ &\text{According to current division rule,} \\ &\qquad i_1 = \frac{R_{eq}}{4000} \times (30 \times 10^{-3}) \\ &\qquad \Rightarrow i_{1} = \frac{2000}{4000} \times (30 \times 10^{-3}) \\ &\qquad \therefore i_1 = 0.015 \ A \end{aligned} \small \begin{aligned} &\qquad i_2 = \frac{R_{eq}}{20000} \times (30 \times 10^{-3}) \\ &\qquad \Rightarrow i_{2} = \frac{2000}{20000} \times (30 \times 10^{-3}) \\ &\qquad \therefore i_2 = 0.003 \ A \\ &\qquad i_3 = \frac{R_{eq}}{5000} \times (30 \times 10^{-3}) \\ &\qquad \Rightarrow i_{3} = \frac{2000}{5000} \times (30 \times 10^{-3}) \\ &\qquad \therefore i_3 = 0.012 \ A \\ &v_2 = i_2 \times 20000 = 0.003 \times 20000 \\ &\therefore v_2 = 60 \ V \\ &\therefore v_{total} = v_2 = 60 \ V \\ &\text{Applying voltage division rule in Fig. 42,} \\ &\qquad v_1 = \frac{3}{3+1} \times v_{total} = \frac{3}{4} \times 60 = 45 \ V \\ &\qquad \therefore p_{3k\Omega} = \frac{(v_1)^2}{3000} = \frac{(45)^2}{3000} = 0.675 \ W \\ &\qquad \therefore p_{20k\Omega} = v_2 \times i_2 = 60 \times 0.003 = 0.18 \ W \end{aligned} \small \begin{aligned} &\text{Therefore,} \\ &\quad v_1 = 45 \ V \text{ and } v_2 = 60 \ V \\ &\quad \scriptsize{\text{the power dissipated in the } 3 \ k\Omega: \ p_{3k\Omega} = 0.675 \ W} \\ &\quad \scriptsize{\text{the power dissipated in the } 20 \ k\Omega: \ p_{20k\Omega} = 0.18 \ W} \\ &\quad \scriptsize{\text{the power supplied by the current source : } p = 1.8 \ W} \end{aligned}Wye-Delta Transformations
Y – Circuit:
Figure 44 – Two forms of the same network: (a) Y , (b) T
\Delta – Circuit:
Figure 45 – Two forms of the same network: (a) \Delta , (b) \Pi
Conversion Method:
Figure 46 – Superposition of Y and \Delta networks as an aid in transforming one to another
Conversion from Delta (\Delta) to Wye (Y) :
\small \begin{aligned} R_1 &= \dfrac{R_b R_c}{R_a + R_b + R_c} \\ \\ R_2 &= \dfrac{R_c R_a}{R_a + R_b + R_c} \\ \\ R_3 &= \dfrac{R_a R_b}{R_a + R_b + R_c} \end{aligned}Conversion from Wye (Y) to Delta (\Delta) :
\small \begin{aligned} R_a &= \dfrac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1} \\ \\ R_b &= \dfrac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2} \\ \\ R_c &= \dfrac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} \end{aligned}The Y and \Delta are said to be balanced when:
\small \begin{aligned} R_1 = R_2 = R_3 = R_Y \quad ; \quad R_a = R_b = R_c = R_\Delta \end{aligned}Under these conditions,
\small \begin{aligned} R_Y = \frac{R_\Delta}{3} \qquad \text{or} \qquad R_\Delta = 3R_Y \end{aligned}Solved Problems on Wye-Delta Transformations
Problem 27
Pb-27: Convert the \Delta network in Fig. 47 to an equivalent Y network.
Figure 47 – Circuit diagram for Pb-27
Solution:
Converting the \Delta to Y ,
Figure 48 – Circuit diagram for solving the Pb-27
\small \begin{aligned} R_1 &= \frac{R_b R_c}{R_a + R_b + R_c} = \frac{10 \times 25}{15 + 10 + 25} = 5 \ \Omega \\ \\ R_2 &= \frac{R_c R_a}{R_a + R_b + R_c} = \frac{25 \times 15}{15 + 10 + 25} = 7.5 \ \Omega \\ \\ R_3 &= \frac{R_a R_b}{R_a + R_b + R_c} = \frac{10 \times 15}{15 + 10 + 25} = 3 \ \Omega \end{aligned}Problem 28
Pb-28: Transform the wye network in Fig. 49 to a delta network.
Figure 49 – Circuit diagram for Pb-28
Solution:
Converting the Y to \Delta ,
Figure 50 – Circuit diagram for solving the Pb-28
\tiny \begin{aligned} R_a &= \dfrac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1} = \dfrac{(10 \times 20) + (20 \times 40) + (40 \times 10)}{10} = 140 \ \Omega \\ \\ R_b &= \dfrac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2} = \dfrac{(10 \times 20) + (20 \times 40) + (40 \times 10)}{20} = 70 \ \Omega \\ \\ R_c &= \dfrac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} = \dfrac{(10 \times 20) + (20 \times 40) + (40 \times 10)}{40} = 35 \ \Omega \end{aligned}Problem 29
Pb-29: Obtain the equivalent resistance R_{ab} for the circuit in Fig. 51 and use it to find current i .
Figure 51 – Circuit diagram for Pb-29
Solution:
Simplifying the circuit,
Figure 52 – Circuit diagram for solving the Pb-29
From the Fig. 52 (b),
\small \begin{aligned} &R_{ab} = ((((1.875 + 12.5)||(2.5 + 10)) + 7.5)||30) \\ &\scriptsize \Rightarrow R_{ab} = \left( \Big( \big( (1.875 + 12.5)^{-1} + (2.5 + 10)^{-1} \big)^{-1} + 7.5 \Big)^{-1} \right. \\ &\qquad \qquad \left. \vphantom{\Big(} + 30^{-1} \right)^{-1} \\ &\therefore R_{ab} = 9.632 \ \Omega \end{aligned} \small \begin{aligned} \therefore i = \frac{120}{R_{ab}} = \frac{120}{9.632} = 12.46 \ A \end{aligned}Problem 30
Pb-30: For the bridge network in Fig. 53, Find R_{ab} and i .
Figure 53 – Circuit diagram for Pb-30
Solution:
Simplifying the circuit,
Figure 54 – Circuit diagram for solving the Pb-30
From the Fig. 54 (b),
\small \begin{aligned} &R_{ab} = ((24 + 6)||(10 + 10)) + 15 + 13 \\ &\scriptsize{\Rightarrow R_{ab} = \Big( (24 + 6)^{-1} + (10 + 10)^{-1} \Big)^{-1} + 15 + 13} \\ &\therefore R_{ab} = 40 \ \Omega \end{aligned} \small \begin{aligned} \therefore i = \frac{240}{R_{ab}} = \frac{240}{40} = 6 \ A \end{aligned}Congratulations! You’ve successfully navigated the fundamental laws of Electrical Circuits. This is a major step toward total mastery of circuit analysis. If any of these concepts, from Ohm’s Law and Kirchhoff’s Laws to Wye-Delta Transformations, still feel a bit confusing, don’t hesitate to reach out. Drop a comment below with your questions, and let’s clear up the confusion together!
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