Understanding Physics Topics is essential for solving complex problems in many fields, including engineering and medicine.
What do you Mean by Node Analysis and Mesh Analysis? What is KVL State Kirchhoffs law?
Kirchhoff formulated the following two laws which enable us to find the distribution of current in complicated electrical circuits (or network of conductors).
Kirchhoff’s First Law or Kirchhoff’s Current Law (KCL)
Statement: In an electrical circuit (or network of wires) the algebraic sum of currents through the conductors meeting at a point is zero, i.e., Σi = 0.
Explanation of the law: Let us consider a number of wires connected at a point A [Fig.]. Currents i1, i2, i3 and i4 flow through these wires in the directions as shown in the Fig.
Here currents i1 and i2 are approaching the point A while currents i3 and i4 are leaving the point A . Taking current entering point A as positive current while current leaving point A as negative current and applying the first law we can write,
i1 + i2 – i3 – i4 = 0 …. (1)
Discussion:
i) Node analysis: Any connecting point in an electrical circuit is called a node. By applying the first law of Kirchhoff, the analysis of an electrical circuit is called node analysis.
ii) Conservation of electric charge: If we write equation (1) in the form i.e., i1 + i2 = i3 + i4, we can say that sum of the currents approaching a connecting point = sum of the currents leaving the point. If the current flows for time t we have,
i1t + i2t = i3t + i4t or, q1 + q2 = q3 + q4
i.e., the sum of the charges approaching the connecting point = sum of the charges leaving the point. The signifi-cance of the above equation is discussed below:
- There cannot be any accumulation of charge at any connecting point in an electrical circuit.
- Charge cannot be created or destroyed, i.e., total charge will remain constant.
Kirchhoff’s Second Law or Kirchhoff’s Voltage Law (KVL)
Statement: The algebraic sum of the product of the current and resistance in any closed loop of a circuit is equal to the algebraic sum of electromotive force acting in that loop, i.e., Σi = Σe.
Explanation of the law: Fig. shows a closed loop ACBDA within an electrical circuit. In the loop, current i1 is in clockwise direction while current i2 is in anticlockwise direction. Taking clockwise current as positive and anticlockwise current as negative, i1 becomes positive and i2 becomes negative.
Again, electromotive force of the sources which send currents in clockwise and anticlockwise direction in the closed loop are taken as positive and negative respectively. So emf e1 becomes negative and emf e2 becomes positive. Hence, for the closed loop ACBDA,
i1r1 – i2r2 = -e1 + e2 ….. (2)
We can write by applying Kirchhoff’s first law to the point A in the Fig.,
i – i1 – i2 = 0 or, i2 = i – i1
So equation (2) can be written as
i1r1 – (i – i1)r2 = -e1 + e2
or, i1(r1 + r2) = e2 – e1 + ir2
or, i1 = \(\frac{\left(e_2-e_1\right)+i r_2}{r_1+r_2}\) ….. (3)
Knowing the values of the quantities of equation (3) we can determine i1 and i2.
Discussion:
i) Mesh analysis: A complicated circuit formed by a number of adjacent loops resembles a mesh which can be analysed one by one by applying Kirchhoff’s second law.
ii) Conservation of energy: We know that inside a source of electricity, other forms of energy are converted into electrical energy and in an external circuit, electrical energy is converted into other forms of energy. The amount of electrical energy developed inside a source of electricity to send one coulomb charge in a circuit is the electromotive force of the source. Again the amount of electrical energy spent in an external circuit due to the flow of one coulomb charge is the potential difference of the external circuit. According to the second law of Kirchhoff,
Σir = Σe i.e., Σe = Σv [potential difference, v = ir]
i.e., electrical energy developed in any circuit due to unit charge = electrical energy spent in the circuit.
So, Kirchhoff’s second law obeys the law of conservation of energy.
Numerical Examples
Example 1.
Two cells, one of emf 1.2 V and Internal resistance 0.5 Ω, the other of emf 2 V and internal resistance 0.1 Ω are connected in parallel and the combination is connected in series with an external resistance of 5 Ω. What is the current through this resistor? [WBJEE 2000]
Solution:
Applying Kirchhoff’s first law to the point A [Fig.] we have,
I1 + I2 = 0 or, I2 = I – I1 ……… (1)
Applying Kirchhoff’s second law to the ioop ACBDA we have,
-I1 × 0.5 + I2 × 0.1 = -1.2 + 2
or, -I1 × 0.5 + (I – I1) × 0.1 = 0.8
or 0.1 I – 0.6I1 = 0.8 ….. (2)
For the loop ADBFA,
-I2 × 0.11 × 5 = -2
or, (I – I,sub>1) × 0.1 + 5I = 2 or, 5.1I – 0.1I1 = 2 …. (3)
Now, multiplying equation (3) by 6 and subtracting equation (2) from it we have,
30.5l = 11.2 or, I = \(\frac{11.2}{30.5}\)= 0.367 A.
Example 2.
Determine the current flowing through the resistor of resistance 200 Ω shown in the circuit diagram Fig. and potential difference across its ends.
Solution:
Applying Kirchhoff’s first law to the point B [Fig.]
we have,
I – I1 – I2 = 0 ….. (1)
Applying Kirchhoff’s second law to the loop ABCA we have,
100I1 + 200I = 10 or, 10I1 + 20I = 1 …. (2)
For the loop ADBA,
-200I — 150I2 = -30
or, 200I + 150(1 – I1) = 30
or, 350I – 150I1 = 30 or, -15I1 + 35I = 3 …. (3)
From the equations (2) and (3) we get,
130I = 9 or I = \(\frac{9}{130}\)A = 69.2 mA
VAB = \(\frac{9}{130}\) × 200 = \(\frac{180}{13}\)V = 13.85 V
Example 3.
Twelve equal wires, each of resistance r ohm, are con-nected so as to form a frame of a cube. An electric current enters this cube at one corner and leaves from the diagonally opposite corner. Calculate the total resistance between the two corners. [J&K CET(Med) ’04]
Solution:
Let ADGCHEFB be the frame of the cube [Fig.] formed by twelve equal wires each of resistance r. Let the total current entering at the corner A and leaving the diagonally opposite corner B be 6x.
Therefore, the 12 wires of the cubic framework (mesh) will have equivalent resistance,
R = \(\frac{V}{I}\) = \(\frac{V}{6 x}\) … (1)
As the resistance of each wire is the same, the current 6x is divided at A into three equal parts, one along AD, the other along AE and the third along AC. At points D, E and C the current is again divided into two equal parts. At points F, G and H the currents combine so that the current in each of the arms FB, GB and HB is 2x. The currents at point B again combine. If V is the potential difference across A and B, then taking the path ACFtBA and applying Kirchhoff’s second law we get,
2x ᐧ r + x ᐧ r + 2x ᐧ r = V
or, 5xr = V …. (2)
From equations (1) and (2) we have,
6xR = 5xr or, R = \(\frac{5}{6} r\)
Node Voltage (Loop Current)
Node voltage: During the analysis of any circuit we can consider a potential for every connecting point i.e., node. This is called node voltage. For analysis of a circuit with the help of node voltage the following rules are adopted:
i) The potential of any one connecting point of the circuit may be taken as zero, because for calculation of current the potential difference between two points is required whatever may be the individual potentials of the two points. In Fig. the potential of C is taken as zero.
ii) If there is no source of electricity or resistance (or condenser or inductance) between any two points, the potentials of the two points are equal.
iii) The potential of the negative terminal of a source of elec-tricity is less than that of the positive terminal by the amount equal to the emf of the source. For example, in the Fig., potential of A = V; potential of B = (V- 5) V.
iv) Current through any resistance is determined by applying the formula \(\frac{V}{R}\) = I. For example, in the Fig., current through BC,
I = \(\frac{V_B-V_C}{10}\) = \(\frac{(V-5)-0}{10}\) = \(\frac{V-5}{10}\)
v) After calculating the currents, Kirchhoff’s first law can be applied at various connecting points.
Loop current: During the analysis of any circuit a separate current for every closed loop may be considered. This is called loop current. For analysis of a circuit with the help of loop current the following rules are adopted:
- Loop current is to be shown with definite direction. For example, in Fig., current i1 for the loop ABCD is shown in the clockwise direction.
- For calculation of current flowing in every branch of a loop the current in the adjacent loop is to be taken into account. For example, in the Fig., current through AB, CD and DA = i1, but current through BC = i1 – i2.
- Next Kirchhoff’s second law can be applied to the loop.
Numerical Examples
Example 1.
Determine the current flowing through the resistance of 5 Ω as shown in the circuit diagram in Fig. (Please note that the solution of the problem has been shown in a different way in the beginning of this chapter.)
Solution:
Potentials of different connecting points are taken in the following way:
VA = V; VB = V — 1.2; VC = V – 2 and VD = 0.
Current along BD = \(\frac{(V-1.2)-0}{0.5}\) = \(\frac{V-1.2}{0.5} \mathrm{~A}\)
Current along CD = \(\frac{(V-2)-0}{0.1}\) = \(\frac{V-2}{0.1} \mathrm{~A}\)
Current along AD = \(\frac{V-0}{5}\) = \(\frac{V}{5} \mathrm{~A}\)
Applying Kirchhoff’s first law to the connecting point D we have,
\(\frac{V-1.2}{0.5}\) + \(\frac{V-2}{0.1}\) + \(\frac{V}{5}\) = 0
or, \(\frac{10(V-1.2)+50(V-2)+V}{5}\) = 0
or, 10V – 12 + 50V – 100 + V = 0
or, 61V = 112 or V = \(\frac{112}{61} \mathrm{~V}\)
∴ Current through resistance of 5 Ω = \(\frac{V}{5}\) = \(\frac{112}{61 \times 5}\) = 0.367 A
Example 2.
In Fig. the resistances of the ammeter and the volt meter are 10 Ω and 900 Ω respectively. What are the readings of the ammeter and the voltmeter?
Solution:
In the circuit, currents flowing in the three loops have been shown as i1, i2 and i3 (clockwise).
From the second law of Kirchhoff we get, for the first loop,
100i1 + 100(i1 – i2) = 0
or, 2i1 – i2 = 0 ….. (1)
for the second loop,
100(i2 – i1) + 10i2 + 100(i2 – i3) = 12
or, -100i1 + 210i2 – 100i3 = 12 ….. (2)
for the third loop,
100(i3 – i2 + 900i3 = 0
or, 10i3 – i2 = 0 … (3)
From (1) and (3)
i2 = 2i1 = 10i3
∴ i1 = 5i3
Substituting these values in equation (2) we get,
-100 × 5i3 + 210 × 10i3 – 100i3 = 12
or, 1500i3 = 12
or, i3 = \(\frac{12}{1500}\) = \(\frac{4}{500} \mathrm{~A}\)
∴ Reading of voltmeter = i3 × 900 = \(\frac{4}{500}\) × 900 = 7.2V
and reading of ammeter
= i2 = 10i3 = 10 × \(\frac{4}{500}\) = 0.08A