3.13 Kirchhoff’s Rules

Many electric circuits cannot be reduced to simple series-parallel combinations. Kirchhoff’s laws (or rules) are used to solve these complicated electric circuits. These rules are basically the expressions of conservation of electric charge and of energy. e.g. two
circuits that cannot be broken down are shown in Figure below.

The two Kirchhoff’s laws are described below.

• Kirchhoff’s Current Law (KCL).
• Kirchhoff’s Voltage Law (KCL).

The two terms related to Kirchhoff’s laws are described below.

Nodes (Junction Point)

A junction in a circuit is a point, where three or more conductors meet. Junctions are also called nodes or branch points. For example, in Fig. (a) points \(D\) and \(C\) are junctions. Similarly, in Fig. (b) points \(B\) and \(F\) are junctions.

Loop

A loop is any closed conducting path. For example, in Fig (a) \(A B C D A, D C E F D\) and \(A B E F A\) are loops. Similarly, in Fig. (b), \(C B F E C, B D G F B\) are loops.

Kirchhoff’s Current Law (KCL).

The algebraic sum of the currents meeting at a point or at a junction in an electric circuit is always zero or the total currents entering a junction equals the total current leaving the junction.
i.e.\(\sum_{\text {junction }} i=0\)
This law can also be written as, “the sum of all the currents directed towards a point (node) in a circuit is equal to the sum of all the currents directed away from that point (node)”. This law is also known as Kirchhoff’s Current Law (KCL).

Thus, in Fig. above according to KCL, \(I_1+I_2=I_3+I_4\). The junction rule is based on the conservation of electric charge.

Example 65: Find the relation in between current \(i_1, i_2, i_3, i_4, i_5\) and \(i_6\).

Solution: \(i_1+i_2-i_3-i_4+i_5+i_6=0\)

Example 66: Find the current in each wire

Solution:

Let potential at point \(B=0\). The potential at other points are mentioned.
\(\therefore\) Potential at E is not known numerically.
Let potential at \(E = x\)
Now applying Kirchhoff’s current law at junction E. (This can be applied at any other junction also).
\(
\frac{x-10}{1}+\frac{x-30}{2}+\frac{x+14}{2}=0 \Rightarrow 4 x =36 \Rightarrow x =9
\)
Current in \(EF =\frac{10-9}{1}=1 A\) from F to E
Current in \(BE =\frac{30-9}{2}=10.5 A\) from B to E
Current in \(DE =\frac{9-(-14)}{2}=11.5 A\) from E to D

Example 67: Find the potential at point \(A\)

Solution: Let potential at \(A = x\), applying Kirchhoff current law at junction A
\(
\frac{x-20-10}{1}+\frac{x-15-20}{2}+\frac{x+45}{2}+\frac{x+30}{1}=0 \Rightarrow \frac{2 x-60+x-35+x+45+2 x+60}{2}=0
\)
\(
\Rightarrow \quad 6 x+10=0 \quad \Rightarrow \quad x=-5 / 3 \quad \Rightarrow \quad \text { Potential at } A=\frac{-5}{3} v
\)

Kirchhoff’s Voltage Law (KVL)

This law states that the algebraic sum of the change in potential around any closed loop involving resistors and cells in the loop is zero. It means that, in any closed part of an electrical circuit, the algebraic sum of the EMFs is equal to the algebraic sum of the products of the resistances and currents flowing through them. It is also known as the loop rule. i.e. \(\sum \Delta V=0\)
This law is also known as Kirchhoff’s Voltage Law (KVL). The loop rule is based on the conservation of energy.

Sign conventions for the application of Kirchhoff’s laws

For the application of Kirchhoff’s laws, the following sign conventions are to be considered.

• The change in potential in traversing a resistance in the direction of current is \(-I R\) while in the opposite direction is \(+I R\).

• The change in potential in traversing an emf source from negative to positive terminal is \(+E\) while in the opposite direction is \(-E\), irrespective of the direction of current in the circuit.    

Example 68: (a) Find currents in different branches of the electric circuit shown in the figure.

(b) Find the potential difference between points \(F\) and \(C\).

Solution: (a) Applying Kirchhoff’s first law ( junction law) at junction \(B\),
\(
i_1=i_2+i_3 \dots(i)
\)
Applying Kirchhoff’s second law in loop 1 (ABEFA),
\(
-4 i_1+4-2 i_1+2=0 \dots(ii)
\)
Applying Kirchhoff’s second law in loop 2 (BCDEB),
\(
-2 i_3-6-4 i_3-4=0 \dots(iii)
\)
Solving Eqs. (i), (ii) and (iii), we get
\(
i_1=1 A
\)

(b) Let us reach from \(F\) to \(C\) via \(A\) and \(B\),
\(
\begin{aligned}
& V_F+2-4 i_1-2 i_3=V_C \\
& V_F-V_C=4 i_1+2 i_3-2
\end{aligned}
\)
\(
\begin{aligned}
& \text { Substituting, } \quad i_1=1 A \\
& \text { and } \quad i_3=-(5 / 3) A \text {, we get } \\
& V_F-V_C=-(4 / 3) V \\
&
\end{aligned}
\)
Here, the negative sign implies that \(V_F<V_C\).

Example 69: (i) Find the potential difference between the points \(A\) and \(B\).

(a) Find the potential difference between the points \(A\) and \(B\).
(b) Find current through 20 V cell, if points \(A\) and \(B\) are connected.

Solution: (i) When we move from \(A\) to \(B\), using KVL,
\(
V_A-9-3 \times 1-8-3 \times 2-7=V_B
\)
\(
V_A-V_B=33 V
\)

No current flows in the branch \(C B\) as \(A B\) is not connected.
Let current in the circuit APSCRQA be \(i\).
Using KVL,
\(
\begin{aligned}
& V_A-i \times 2-20-i \times 2-i \times 1-i \times 2-i \times 2+10-i \times 1=V_A \\
& \Rightarrow 10 i=-10 \Rightarrow i=-1 A
\end{aligned}
\)
i.e., direction of \(i\) is opposite.
Now, \(A\) to \(B\) path will be \(A P S C B\),
\(
\begin{gathered}
V_A-i \times 2-20-i \times 2-i \times 1+5=V_B \\
V_A-V_B=15+5 i=15+5 \times(-1)=10 V
\end{gathered}
\)

(b) Now, if the points \(A\) and \(B\) are connected.

Resistance \(2 \Omega, 2 \Omega\) and \(1 \Omega\) are in series along APSC and resistance \(1 \Omega, 2 \Omega\) and \(2 \Omega\) are also in series along \(A Q R C\).

So, the equivalent voltage will be
\(
E=\frac{\frac{20}{5}+\frac{5}{1}+\frac{10}{5}}{\frac{1}{5}+\frac{1}{1}+\frac{1}{5}}=\frac{4+5+2}{\frac{1+5+1}{5}}=\frac{11 \times 5}{7}=\frac{55}{7} V
\)
\(\therefore\) Potential difference between points \(A\) and \(B\),
\(
V_A-V_B=\frac{55}{7} V
\)
For cell of emf 20 V ,
\(
\begin{aligned}
V_A-V_B & =20-5 i \\
\frac{55}{7} & =20-5 i \Rightarrow 5 i=20-\frac{55}{7} \simeq 12 A
\end{aligned}
\)
\(\therefore\) Current through cell of \(20 V, \dot{i}=2.4 A\)

Example 70: An electric heater and an electric bulb are rated \(500 W, 220 V\) and \(100 W, 220 V\) respectively. Both are connected in series to a 220 V a.c. mains. Calculate the power consumed by (i) heater (ii) bulb.

Solution:
\(P=\frac{V^2}{R}\) or \(R=\frac{V^2}{P}\), For heater, Resistance \(R_h=\frac{(220)^2}{500}=96.8 \Omega\). For bulb, resistance \(R_b=\frac{(220)^2}{100}=484 \Omega\)
Current in the circuit when both are connected in series \(I =\frac{ V }{ R _{ L }+ R _{ h }}=\frac{220}{484+96.8}=0.38 A\)
(i) Power consumed by heater \(=I^2 R _{ h }=(0.38)^2 \times 96.8=13.98 W\)
(ii) Power consumed by bulb \(=I^2 R_b=(0.38)^2 \times 484=69.89 W\)

Example 71: A heater coil is rated \(100 W, 200 V\). It is cut into two identical parts. Both parts are connected together in parallel to the same source of 200 V . Calculate the energy liberated per second in the new combination.

Solution: \(\because P =\frac{ V ^2}{ R } \quad \therefore R =\frac{ V ^2}{ P }=\frac{(200)^2}{100}=400 \Omega\)
Resistance of half piece \(=\frac{400}{2}=200 \Omega\)
Resistance of pieces connected in parallel \(=\frac{200}{2}=100 \Omega\)
Energy liberated/second \(P =\frac{ V ^2}{ R }=\frac{200 \times 200}{100}=400 W\)

Example 71: When a battery sends current through a resistance \(R _1\) for time t the heat produced in the resistor is Q . When the same battery sends current through another resistance \(R_2\) for time \(t\), the heat produced in \(R_2\) is again \(Q\). Determine the internal resistance of the battery.

Solution: \(\left[\frac{E}{R_1+r}\right]^2 R_1=\left[\frac{E}{R_2+r}\right]^2 R_2 \Rightarrow r=\sqrt{R_1 R_2}\)

Example 72: Three \(60 W, 120 V\) light bulbs are connected across a 120 V power source. If the resistance of each bulb does not change with current then find out the total power delivered to the three bulbs.

Solution: 

Total power supplied \(=\frac{ V ^2}{\left(\frac{3}{2}\right) R }=\left(\frac{2}{3}\right)\left(\frac{ V ^2}{ R }\right)=\frac{2}{3} \times 60=40 W\)

Example 73: An apparatus is connected to an ideal battery as shown in the figure. For what value of current, power delivered to the apparatus will be maximum?

Solution: For maximum power \(R _{\text {ext }}= R _{ int }= R \therefore\) current \(=\frac{ E }{2 R }\)

Example 74: Three identical bulbs are connected in a circuit as shown. What will happen to the brightness of bulbs A & B if the switch is closed?

Solution: When switch \(S\) is closed, current does not pass through bulb \(C\). Hence total resistance decreases & current passing through bulbs \(A\) & \(B\) increases. Brightness of bulbs \(A\) & \(B\) also increases.

Example 75: Determine the current in each branch of the network shown in Fig. below.

Solution: Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns \(I_1, I_2\) and \(I_3\) which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff’s second rule for the closed loop ADCA gives,
\(
10-4\left(I_1-I_2\right)+2\left(I_2+I_3-I_1\right)-I_1=0 \dots(i)
\)
that is, \(7 I_1-6 I_2-2 I_3=10\)
For the closed-loop ABCA, we get
\(
10-4 I_2-2\left(I_2+I_3\right)-I_1=0
\)
that is, \(I_1+6 I_2+2 I_3=10 \dots(ii)\)
For the closed-loop BCDEB, we get
\(
5-2\left(I_2+I_3\right)-2\left(I_2+I_3-I_1\right)=0
\)
that is, \(2 I_1-4 I_2-4 I_3=-5 \dots(iii)\)
Equations (i), (ii) & (iii) are three simultaneous equations in three unknowns. These can be solved by the usual method to give
\(
I_1=2.5 A, \quad I_2=\frac{5}{8} A, \quad I_3=1 \frac{7}{8} A
\)
The currents in the various branches of the network are
\(
\begin{aligned}
& AB : \frac{5}{8} A, \quad CA : 2 \frac{1}{2} A, \quad DEB : 1 \frac{7}{8} A \\
& AD : 1 \frac{7}{8} A, \quad CD : 0 A, \quad BC : 2 \frac{1}{2} A
\end{aligned}
\)
It is easily verified that Kirchhoff’s second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
\(
5 V+\left(\frac{5}{8} \times 4\right) V -\left(\frac{15}{8} \times 4\right) V
\)
equal to zero, as required by Kirchhoff’s second rule.