Kirchhoff's Laws for current and voltage lie at the heart of circuit analysis. With these two laws, plus the equations for individual component (resistor, capacitor, inductor), we have the basic tool set we need to start analysing circuits.
This article assumes you are familiar with the definitions of node,distributed node,branch and loop.Both of Kirchhoff's laws can be understood as corollaries of Maxwells equations in the low-frequency limit. They are accurate for DC circuits, and for AC circuits at frequencies where the wavelengths of electromagnetic radiation are very large compared to the circuits.
Kirchhoff’s current law:
This law is also called Kirchhoff's first law, Kirchhoff's point rule, or Kirchhoff's junction rule (or nodal rule).
The principle of conservation of electri charge, combined with the very large repulsive Coulomb forces that will occur if charge "piles up" anywhere, imply that:
At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to
the sum of currents flowing out of that node or equivalently
The algebraic sum of currents in a network of conductors meeting at a point is
zero.Recalling that current is a signed (positive or negative) quantity reflecting direction towards
or away from a node, this principle can be stated as:
n is the total number of branches with currents flowing towards or away from the node.
This formula is valid for complex currents:
The law is based on the conservation of charge whereby the charge (measured in coulombs) is the product of the current (in amperes) and the time (in seconds)
Uses:
A matrix version of Kirchhoff's current law is the basis of most circuit simulation software, such as spice.Kirchhoff's current law combined with ohms law is used in nodal analysis.
KCL is applicable to any lumped network irrespective of the nature of the network; whether unilateral or bilateral, active or passive, linear or non-linear.
Kirchhoff’s voltage law:
This law is also called Kirchhoff's second law, Kirchhoff's loop (or mesh) rule, and Kirchhoff's second rule.
The principle of conservation of energy implies that The directed sum of the electrical potential diference (voltage) around any closed network is zero, or:
More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop, or: The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.
Similar to KCL, it can be stated as:
Here, n is the total number of voltages measured. The voltages may also be complex:
This law is based on the conservation of energy whereby voltage is defined as the energy per unit charge. The total amount of energy gained per unit charge must be equal to the amount of energy lost per unit charge, as energy and charge are both conserved.
Limitations:
KCL and KVL both depend on the lumped element model being applicable to the circuit in question. When the model is not applicable, the laws do not apply.
KCL, in its usual form, is dependent on the assumption that current flows only in conductors, and that whenever current flows into one end of a conductor it immediately flows out the other end. This is not a safe assumption for high-frequency AC circuits, where the lumped element model is no longer applicable. It is often possible to improve the applicability of KCL by considering "parasitic capacitancespa" distributed along the conductors. Significant violations of KCL can occur even at 60 Hz, which is not a very high frequency.
In other words, KCL is valid only if the total electric charge, Q remains constant in the region being considered. In practical cases this is always so when KCL is applied at a geometric point. When investigating a finite region, however, it is possible that the chrge density within the region may change. Since charge is conserved, this can only come about by a flow of charge across the region boundary. This flow represents a net current, and KCL is violated.
KVL is based on the assumption that there is no fluctuating magnetic field linking the closed loop. This is not a safe assumption for high-frequency (short-wavelength) AC circuits.In the presence of a changing magnetic field the electric field is not a conservative vector field. Therefore, the electric field cannot be the gradient of any potential. That is to say, the line integral of the electric field around the loop is not zero, directly contradicting KVL.
It is often possible to improve the applicability of KVL by considering "parasitic inductances" (includingmutual inductances) distributed along the conductors. These are treated as imaginary circuit elements that produce a voltage drop equal to the rate-of-change of the flux.
Conventions for Loop Rule:
• The following conventions apply for determining the sign of delta V across circuit elements. The travel direction is the direction that we choose to proceed around the loop
Procedure for Applying Rules
1. Assume all voltage sources and resistances are given. (If not label them V1, V2 ..., R1, R2 etc)
2. Label each branch with a branch current. (I1, I2, I3 etc)
3. Apply junction rule at each node.
4. Applying the loop rule for each of the independent loops of the circuit. 5. Solve the equations by substitutions/linear manipulation.
1.Kirchoff’s law of meshes is in accordance with law of conservation of
Solution:
Kirchhoff’s law of meshes is in accordance with law of conservation of energy.
2.Kirchoff’s law of junctions is also called the law of conservation of?
Solution:
Kirchhoff’s law of junctions is also called the law of conservation of charge.
1.Six resistors of each 2 ohm are connected as shown in the figure. The resultant resistance between A and B is.
Solution: The resultant resistance between A and B is.1
2.In the given circuit current through the galvanometer is
Solution: current through the galvanometer is Flows from D to C.
1.In figure, find I1, I2 and I3 if switch S is open.
Solution:
a) When S is open, I3 = 0, because no current can flow through the open switch. Applying the KCL the to point a, we get
I1 + I3 = I2 or I1 = I2
Applying the KVL to loop acbda. we get
– 7I1 + 12 – 9 – 8I2 = 0 … (i)
Because I2 = I1, (i) becomes
15I1 = 3 or I1 = 0.20A
Also, I2 = I1 = 0.20A.
Note that this is the same result that one would obtain by replacing the two batteries by a single 3V battery.
2.In the given circuit find the current passing through each branch.
Solution
Kirchhoff’s current law does not allow any current through 1 W resistor of the circuit. Now the circuit will become as shown in figure. Using KVL, \({{I}_{1}}=1\) A, \({{I}_{2}}=1\) A
1.In figure, find I1, I2 and I3 if switch S is closed.
Solution:
With the switch S closed, I3 is no longer known to be zero. Therefore, by applying the KCL at point a gives
I1 + I3 = I2 … (i)
Applying KVL to loop acba.
– 7I1 + 12 + 4I3 = 0 … (ii)
and loop abda gives
– 9 – 8I2 – 4I3 = 0 … (iii)
After solving equation (i), (ii) and (iii) for I1, I2 and I3. From (iii), we get
I3 = –2I2 – 2.25
Substituting this in (ii) gives
+ 12 – 7I1 – 9 – 8I2 = 0 or 7I1 + 8I2 = 3
Also substituting for I3 in (i) gives
I1 – 2I2 – 2.25 = I2 or I1 = 3I2 + 2.25
Substituting this value in the previous equation,
21I2 + 15.75 + 8I2 = 3 or I2 = – 0.44A
Using this in the equation for I1 gives
I1 = 3 (– 0.44) + 2.25 = – 1.32 + 2.25 = 0.93A
Note that the minus sign is a part of the value we have found for I2. It must be carried along with its numerical value. Now we can used (i) to find
I3 = I2 – I1 = (– 0.44) – 0.93 = – 1.37A
Note that this problem could not be solved by simple parallel-or series resistance techniques.
Solution:
The current assumed in the circuit as shown in the figure. Applying KCL to loop abca.
+3 - 3I1 + 6 = 0
or I1 = 3A
Applying KVL to loop abda
+3-6 (I - I1) + 3 = 0 or I - I1 = 1
I = 1 + I1 = 4A
The current through each branch is shown in the figure.
An electrical cell is a device that is used to generate electricity, or one that is used to make chemical reactions possible by applying electricity.
i) A cell is device which by converting chemical energy into electrical energy maintains the flow of charge in a circuit.
ii) The emf of a cell is defined as the work done by the cell in moving unit positive charge in the whole circuit including the cell once.
S.I unit of emf is volt
iii) The emf of a cell depends only on the nature of electrodes and electrolyte.
Every cell has two terminals namely:
There are different types of cells available and some of them are as follows:
Primary Cells:
Voltaic, Leclanche, Daniel and Dry cells are primary cells. They convert chemical energy into electrical energy. They can’t be recharged. They supply small currents.
Secondary Cells (or) Storage Cells:
Ø Electrical energy is first converted into chemical energy and then the stored chemical energy is converted into electrical energy due to these cells.
Ø These cells can be recharged.
Ø the internal resistance of a secondary cell is low whereas the internal resistance of a primary cell is large.
diagram:
Internal resistance:
a) Internal resistance of a cell is the resistance offered by the electrolyte between anode and cathode of the cell.
b) It depends on
(i) area and size of plates (electrodes)
(ii) distance between the plates (electrodes)
(iii) strength of electrolyte
(iv) temperature
c) Internal resistance decreases with increase of temperature.
When there is current present in the device or the electrical circuit and there’s a voltage drop in source voltage or source battery is internal resistance. It is caused due to electrolytic material in batteries or other voltage sources.
Internal Resistance (r) = (E – V)/I
Where E is the emf of the device; V is the potential difference between the device; I is the current in the device. Internal Resistance is the result of the resistance in the battery or the accumulation in the battery. the equation used to derive this is as follows:
V = (E – Ir)
It is the resistance offered by the electrolyte of the cell.
It depends on :
Cells are basically electrical elements. ... Thus, as two 1.2-volt cells are connected in series we get a cell whose net voltage is 2.4 volts. Thus, a series combination of cells can produce higher voltages.
Cells in Series :
Cells are basically electrical elements. Cells employ chemical reaction to produce electrical potential.
Series cells are produced by individual cells connected together. It is done by connecting the positive terminal of a battery to the negative terminal of another.
Series cells are usually used by certain equipments which need higher voltages. In order to produce higher voltages we connect the cells in series.
In this arrangement, the positive terminal of one cell is connected to negative terminal of other in succession. Figure represents (n) cells, each of e.m.f (E) and internal resistance (r) connected in series with an external resistance (R)
i) Since all the cells are in series,net e.m.f = nE.
As all the internal resistances are in series, so net internal resistance = nr
iii) Total resistance of circuit = R + nr
iv) Current I = \(\frac{\text{Net}\,\text{e}\text{.m}\text{.f}}{\text{Net }\,\text{resistance}}\)
\(I\,=\frac{nE}{R+nr}\)
a) If R >> nr, then \(I\,\,=\,\,n\frac{E}{R}\,\)
b) If R <<nr, then \(I\,=\,\frac{E}{r}\).
When net external resistance is far greater than net internal resistance, then to get maximum current, cells must be connected in series
Also if two cells of different emf’s are in series
Eequivalent = E1–E2
requivalent = r1 + r2
i = \(\frac{{{E}_{1}}-{{E}_{2}}}{R+{{r}_{1}}+{{r}_{2}}}\)
First cell is discharging then V1 = E1 – ir1
Second cell is charging then V2 = E2 + ir2
Cell having more emf is in discharging state and cell having less emf in charging state.
Wrongly connected cells in series :
N cells each of emf ‘E’ are to be connected in series. If ‘n’ cells of them are wrongly
connected, then
net emf = (N – 2n)E and
net internal resistance = Nr.
CELLS IN PARALLEL:
When positive terminals of every cell are connected together, and negative cells are connected in a cell, then we call the cells as parallel cells. These cells are usually used by devices which need to produce higher current. In this arrangement, the positive terminals of all cells are connected to one point and negative terminals to one point. Figure represents (m) cells, each of e.m.f (E) and internal resistance (r), connected in parallel and an external resistance R is connected across the combination.
Since all the cells are in parallel, the net e.m.f equals to e.m.f due to a single cell.
Net e.m.f = E
ii) similarly all resistances are in parallel, so
Net internal resistance = \(\frac{r}{m}\)
iii) Total resistance of circuit = \(R+\frac{r}{m}\)
iv) Current I = \(\frac{E}{R+\left( \frac{r}{m} \right)}\)
a) If R << \(\frac{r}{m}\), then \(I=\frac{mE}{r}\)
b) If R >>\(\frac{r}{m}\) , then \(I=\frac{E}{R}\)
when net internal resistance is far greater than net external resistance, then to get maximum current, the cells must be connected in parallel.
If two cells of different emf are connected in parallel then
\({{E}_{eq}}=\frac{\Sigma \,\,\,E/r}{\Sigma \,1/r\,}=\frac{\left( \frac{{{E}_{1}}}{{{r}_{1}}}+\frac{{{E}_{2}}}{{{r}_{2}}} \right)}{\left( \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}} \right)}=\left( \frac{{{E}_{1}}{{r}_{2}}+{{E}_{2}}{{r}_{1}}}{{{r}_{1}}+{{r}_{2}}} \right)\)
\(\frac{1}{{{r}_{eq}}}=\sum{\frac{1}{r}=\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}}\)
\({{r}_{eq}}=\left( \frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}} \right)\)
\(i=\frac{{{E}_{eq}}}{{{r}_{eq}}+R}=\frac{{{E}_{1}}{{r}_{2}}+{{E}_{2}}{{r}_{1}}}{{{r}_{1}}{{r}_{2}}+R\left( {{r}_{1}}+{{r}_{2}} \right)}\)
MIXED GROUPING :
In this arrangement, cells are connected both in series and parallel.
This arrangement comprises of ‘m’ rows of cells in parallel, each row containing ‘n’ cells in series. Total no. of cells = mn.
i) The e.m.f of each cell is (E) and internal resistance is (r).
ii) The combination is connected to external resistance R. Net e.m.f = nE
iii) Net internal resistance = \(\frac{nr}{m}\)
iv) Net resistance of circuit = R + \(\frac{nr}{m}\)
v) Current I = \(\frac{nE}{R+\frac{nr}{m}}=\frac{mnE}{mR+nr}\)
vi) Thus, for maximum current through external resistance, cells should be connected in mixed grouping such that external resistance is equal to net internal resistance i.e.
m R = nr and \({{I }_{\max }}=\frac{mE}{2r}=\frac{nE}{2R}\)
vii) Power transferred to the load will be maximum when external resistance is equal to internal resistance. (mR = nr)
viii) Mixed grouping is preferred when more power is needed.
the difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other. If the electric potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1 joule of potential energy when moved between those two locations. ... Because electric potential difference is expressed in units of volts, it is
sometimes referred to as the voltage.
Two simple circuits and their corresponding electric potential diagrams are shown below.
Terminal Potential difference:
The Terminal Potential difference is defined as the work done in moving unit positive charge from anode to cathode of cell through external resistance. If ‘i’ current flows through the external resistance ‘R’ then V= iR.
Lost voltage:
The voltage dropped across the internal resistance is called lost voltage and is equal ‘ir’.
Relation among E,V and r
When current is drawn from a cell, current inside the cell is from cathode to anode and current outside cell is from anode to cathode. Then terminal p.d, V = E - ir. i.e V < E.
When cell is charging
When current is sent into a cell it is said to be in charging condition.
a) Current inside cell is from anode to cathode
b) When a cell is being charged potential difference across its terminals is greater than emf of the cell. Then terminal p.d is
V=E+ir i.e V > E
A cell of emf E and internal resistance r are connected to external resistance R then
a) If R0and R ,the circuit is said to be closed then
i ==\(\frac{E}{R+r}\)\(\frac{{{E}_{effective}}}{{{R}_{effective}}}\)
EMF of a Cell:
When a cell of EMF ‘E’ and internal resistance ‘r’ is connected to an external
resistance ‘R’ as shown, where i = current in the circuit. Now the cell discharges or
delivers current i through R. Then,
V = potential difference across the external resistance
V1 = Voltage across internal resistance (or) lost volts,
EMF of the cell, E = V + V1 = V + ir
Ø From Ohm’s law, i =\(\frac{E}{\left( R+r \right)}=\frac{V+{{V}^{'}}}{\left( R+r \right)}\)
Ø V = iR = \(\frac{ER}{\left( R+r \right)}\)
Ø Fractional energy useful =\(\frac{V}{E}=\frac{R}{R+r}\)
Ø % of fractional useful energy \(\left( \frac{V}{E} \right)\,\,\,100\)=\(\left( \frac{R}{R+r} \right)100\)
Ø Fractional energy lost, \(\frac{{{V}^{'}}}{E}=\frac{r}{R+r}\)
Ø % of lost energy,\(\frac{{{V}^{'}}}{E}=\frac{r}{R+r}\).
Ø internal resistance, r = \(\left[ \frac{E-V}{V} \right]R\).
Difference Between emf and Potential Difference:
Back emf :
It is the emf developed in the cell against the emf of the cell due to the ionic polarization.
i) To prevent the polarisation in Voltaic cell and Leclanche cell, the depolarisers like potassium dichromate (or) manganese dioxide is to be added to electrolyte.
ii) Depolariser is essentially an oxidising agent that neutralises hydrogen and there by prevents the covering of the copper plate and nullifies back emf.
1.The direction of current in a cell is?
Solution:
The direction of current in a cell is \( \left( - \right)\)ve pole to \(\left( + \right)\) ve pole during discharging.
2.Out of the statements :
(A) The potential difference across battery may be equal to its emf
(B) The potential difference across battery may be greater than its emf
(C) The potential difference across battery may be less than its emf
1) A and B are correct, C is wrong
2) A is correct, B and C are wrong
3) B is correct, A and C are worn
4) A,B and C are correct
Solution:
A,B and C are correct
(A) The potential difference across battery may be equal to its emf
(B) The potential difference across battery may be greater than its emf
(C) The potential difference across battery may be less than its emf.
1. A 9.0 V battery has an internal resistance of 12.0 Ω. (a) What is the potential difference across its terminals when it is supplying a current of 50.0 mA?
(b)What is the maximum current this battery could supply?
Solution:
(a) pd = E – I r = 9 – (50 x 10-3 x 12) = 8.4 V
(b) Max current = E/r = 9 / 12 = 0.75 A.
2.A battery is connected in series with a variable resistor and an ammeter. When the resistance of the resistor is 10 Ω the current is 2.0 A. When the resistance is 5 Ω the current is 3.8 A. Find the emf and the internal resistance of the battery.
Solution:
E = I(R +r) E = 2 (10 + r) and
E = 3.8 (5 + r) so r= 0.56 Ω E = 20 + (2 x 0.56) = 21.1 V.
1.What is the terminal p.d. for a cell of emf 2V and internal resistance 1 ohm when it is connected to a 9 ohm resistor?
Solution:
Just pretend the internal resistance is one of the normal resistors in the circuit. Draw it in the circuit diagram next to the cell so that all the current that goes through the cell also goes through the resistor.
To find V, the terminal pd (or the voltage available to the external circuit), calculate the current, I, for the whole circuit:
Note: VT and RT are the voltage and resistance for the whole circuit, including external and internal resistance.
Therefore, the 9Ω resistor gets V = IR = 0.2 x 9 = 1.8V
So this 2V emf cell actually supplies 1.8V to the external circuit.
2.A ‘potato’ cell has emf 1.0 V and internal resistance 5000 Ω. Roughly how many of these cells in what arrangement would adequately light a 5 W, 6.0 V filament lamp?
Solution:
Required pd = 6 V and power = 5 W so current required = P/V = 5/6 = 0.83
A 6 cells in series have an internal resistance of 6 x 5000 = 30000 Ω . If the total
internal resistance is low, say 0.1 Ω then 6 - 0.083 = 5.92 V
roughly which would light the lamp well, so 30,000/n = 0.1 and n = number of
combinations = 300,000
so total number of cells is 6 x 300,000 = 1,800,000.
1.In a series grouping of N cells current in the external circuit is I. How may cells should be reversed their polarity that the current becomes \(\frac{1}{3}rd\) the earlier value.
Solution :
Before reversing the cells current is
I =\(\frac{NE}{R+Nr}\)
Let n cells be reversed in their polarity
Net e.m.f. = (N – n)E – nE
= (N-2n) E
Total resistance Nr + R
\(\Rightarrow {i}'\,\,=\,\,\frac{\left( N-2n \right)\,E}{Nr+R}\)
But, \(\frac{{{I}'}}{I}\,\,=\,\,\frac{1}{3}\)=\(\frac{\left( N-2n \right)\,E}{Nr+R}\)
Þ n =\(\frac{N}{3}\)
2.In figure A B is 300 cm long wire having resistance 10W per meter. Rheostat is set at 20W. The balance point will be attained
at
Solution:
\({{V}_{AB}}=\frac{6\times 30}{50}=3.6V.\)
Terminal voltage of cell = \(\frac{2\times 1.5}{2}=1.5V\)
Using \(V=kl\,\,\Rightarrow \,\,1.5=\frac{3.6}{300}\)l
or l = 125 cm.