Kirchhoff’s Circuit Laws - KCL & KVL | Symmetric line method

We have studies Ohm's law but in real life we have to deal with a lot complex circuits. To calculate current and potential differences across them we require something else. So here we will know about Kirchhoff's circuit laws. Contents:

1. Kirchhoff's circuit laws
2. Kirchhoff’s First Law or Kirchhoff’s Current Law
3. Kirchhoff’s Second Law or Kirchhoff’s Voltage Law
4. Dissimilar cells in Parallel
5. Symmetric line method

Kirchhoff’s Circuit Laws

Gustav Kirchhoff, a German physicist, in 1825 developed a set of rules or laws to find the voltages and currents circulating in complex circuits where working only with Ohm's law don't suffice. There are majorly two laws to calculate current and potential difference along different loops present in a circuit known as Kirchhoff's circuit laws. For the calculations, to obtain circuit equations these laws are useful.

The first rule, the junction theorem, states that the sum of the currents into a specific junction in the circuit equals the sum of the currents out of the same junction. Electric charge is conserved: it does not suddenly appear or disappear; it does not pile up at one point and thin out at another.

The second rule, the loop equation, states that around each loop in an electric circuit the sum of the emf’s (electromotive forces, or voltages, of energy sources such as batteries and generators) is equal to the sum of the potential drops, or voltages across each of the resistances, in the same loop. All the energy imparted by the energy sources to the charged particles that carry the current is just equivalent to that lost by the charge carriers in useful work and heat dissipation around each loop of the circuit. [source]

Kirchhoff’s First Law or Kirchhoff’s Current Law (KCL)

In any electrical network, the algebraic sum of currents meeting at a junction is always zero i.e. ∑I = 0

This law is also called as Junction rule.

$I_{1}-I_{2}+I_{3}+I_{4}-I_{5}=0$

or $\mathrm{I}_{1}+\mathrm{I}_{3}+\mathrm{I}_{4}=\mathrm{I}_{2}+\mathrm{I}_{5}$

Kirchhoff’s Second Law or Kirchhoff’s Voltage Law (KVL)

The algebraic sum of all the potential drops and emf’s along any closed path in a network is zero. i.e. ∑V = 0

This law is also called as Loop rule.

Dissimilar cells in Parallel

Let two cells of emf ${E_1}$ and ${E_2}$ with internal resistances ${r_1}$ and ${r_2}$ be connected in parallel to an external resistance$R$.

Applying loop rule to loop ${\rm{d}}$${\rm{l}}$${\rm{m}}$${\rm{n}}$${\rm{c}}$${\rm{d}}$ $ - {\rm{IR}} - {{\rm{I}}_1}{{\rm{r}}_1} + {{\rm{E}}_1} = 0\quad \ldots \ldots $ (1)

Applying loop rule to loop ${\rm{a}}$${\rm{l}}$${\rm{m}}$${\rm{n}}$${\rm{b}}$${\rm{a}}$ $ - {\rm{IR}} - {{\rm{I}}_2}{{\rm{r}}_2} + {{\rm{E}}_2} = 0\quad \ldots \ldots $ (2)

Applying first rule at junction $\ell {\rm{I}} = {{\rm{I}}_1} + {{\rm{I}}_2}$

Multiply equation 1 by ${r_2}$ and eqn. 2 by ${r_1}$ and put ${I_2} = I - {I_1}$ we get

$ - {\rm{IR}}{{\rm{r}}_2} - {{\rm{I}}_1}{{\rm{r}}_1}{{\rm{r}}_2} + {{\rm{E}}_1}{{\rm{r}}_2} = 0\quad $ and $\quad - {\rm{IR}}{{\rm{r}}_1} - \left( {{\rm{I}} - {{\rm{I}}_4}} \right){{\rm{r}}_2}{{\rm{r}}_1} + {{\rm{E}}_2}{{\rm{r}}_1} = 0$

[caption id="attachment_4627" align="aligncenter" width="222"] Dissimilar cells in Parallel[/caption]
On adding these we get $\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}=\mathrm{IR}\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)+\mathrm{Ir}_{1} \mathrm{r}_{2}$

$=\mathrm{I}\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)\left[\mathrm{R}+\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right]$

Or, $\quad {\rm{I}} = {{\left( {{{\rm{E}}_{{{\rm{f}}_2}}} + {{\rm{E}}_2}{{\rm{r}}_1}} \right)/{{\rm{r}}_1} + {{\rm{r}}_2}} \over {{\rm{R}} + {{{{\rm{r}}_1}{{\rm{r}}_2}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}}}} = {{{{\rm{E}}_{{\rm{eq}}}}} \over {{\rm{R}} + {{\rm{r}}_{{\rm{eq}}}}}}$

[caption id="attachment_4628" align="aligncenter" width="198"] Equivalent Cell[/caption] Two dissimilar cells in parallel are equivalent to a single cell of internal resistance ${{\rm{r}}_{{\rm{eq}}}} = {{{{\rm{r}}_{\rm{r}}} \cdot {{\rm{r}}_2}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}}$

and ${\rm{emf}} {{\rm{E}}_{{\rm{eq}}}} = {{{{\rm{E}}_1}{{\rm{r}}_2} + {{\rm{E}}_2}{{\rm{r}}_1}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}} = {{{{\rm{r}}_1}{{\rm{r}}_2}} \over {{{\rm{r}}_1} + {{\rm{r}}_2}}}\left[ {{{{{\rm{E}}_1}} \over {{{\rm{r}}_1}}} + {{{{\rm{E}}_2}} \over {{{\rm{r}}_2}}}} \right]$

Symmetric line method

If a circuit is symmetric about a line then all points lying on line of symmetry are at same potential.

In this method, we divide the circuit in two equal parts and double the resistance of one part.

Example: In above question DF is line of symmetry so D and F are at same potential.

Now ${{\rm{R}} \over 2}$ is in parallel to ${\rm{R}}({\rm{CD}}) = {{{\rm{R}} \times {\rm{R}}/2} \over {{\rm{R}} + {\rm{R}}/2}} = {{\rm{R}} \over 3}$

This ${{\rm{R}} \over 3}$ is in series with ${\rm{R}}({\rm{AC}}) = {\rm{R}} + {\rm{R}}/3 = 4{\rm{R}}/3$

This ${{\rm{4R}} \over 3}$ is in parallel with ${\rm{R}}({\rm{AD}}) = 4{\rm{R}}/7\quad $

So ${{\rm{R}}_{{\rm{eq}}}} = {{2 \times 4{\rm{R}}} \over 7} = {{8{\rm{R}}} \over 7}$

Solved Examples

[esquestion]A circuit has a section ABC. If the potential at points $A, B$ and $C$ are ${V_1},{V_2}\&\,\,{V_3}$ then calculate potential at O. #sol#Applying junction rule at O $\quad - {I_1} - {I_2} - {I_3} = 0$ or $\quad {I_1} + {I_2} + {I_3} = 0$ Let ${V_0}$ be potential at O then ${{\rm{V}}_0} - {{\rm{V}}_1} = {{\rm{I}}_1}{{\rm{R}}_1},\quad {{\rm{V}}_0} - {{\rm{V}}_2} = {{\rm{I}}_2}{{\rm{R}}_2},\quad {{\rm{V}}_0} - {{\rm{V}}_3} = {{\rm{I}}_3}{{\rm{R}}_3}$ so $\quad {{\rm{I}}_1} = {{{{\rm{V}}_0} - {{\rm{V}}_1}} \over {{{\rm{R}}_1}}},\quad {{\rm{I}}_2} = {{{{\rm{V}}_0} - {{\rm{V}}_2}} \over {{{\rm{R}}_2}}},{{\rm{I}}_3} = {{{{\rm{V}}_0} - {{\rm{V}}_3}} \over {{{\rm{R}}_3}}}$ adding them we get $\quad {{\rm{V}}_0}\left[ {{1 \over {{{\rm{R}}_1}}} + {1 \over {{{\rm{R}}_2}}} + {1 \over {{{\rm{R}}_3}}}} \right] - \left[ {{{{{\rm{V}}_1}} \over {{{\rm{R}}_1}}} + {{{{\rm{V}}_2}} \over {{{\rm{R}}_2}}} + {{{{\rm{V}}_3}} \over {{{\rm{R}}_3}}}} \right] = 0$ or $\quad {V_0} = {{{V_1}} \over {{R_1}}} + {{{V_2}} \over {{R_2}}} + {{{V_3}} \over {{R_3}}}/{1 \over {{R_1}}} + {1 \over {{R_2}}} + {1 \over {{R_3}}}$ [/esquestion] Also Read:

• Types and Effects of Electric Current
• Ohm’s Law and Resistance
• Combination of Resistances
• EMF and Internal Resistances of a Cell
• Cells Connected in Series, parallel and Mixed
• Kirchhoff’s Circuit Law
• Electric Currents in Conductors
• Wheatstone Bridge
• Post office Box
• Wheatstone Meter Bridge
• Moving Coil galvanometer
• Ammeter and Voltmeter
• Potentiometer Working Principle