Current Electricity: Overview, Questions, Preparation

Gustav Robert Kirchhoff was a German physicist born in Russia. His work involved researching electrical conduction. It led to him formulating two laws known as Kirchhoff's Current law and Kirchhoff's Voltage law. These laws are generally used for the analysis of circuits. They help in calculating the resistance and the flow of current in different streams through the network.

The topic 'Kirchhoff's Laws' falls under the chapter 'Current Electricity'. It is the second unit of NCERT Class 12 Physics. Along with the first unit 'Electrostatics', this unit comes for 16 marks in the exams.

Sometimes electrical circuits are interconnected in a complicated way. The parallel and series combination of resistors or capacitors are not always sufficient to determine all the currents and potential differences. Hence two laws known as the Kirchhoff's Laws are used for the analysis of electrical circuits. Kirchhoff's Laws are divided as:

• Junction Rule

Also known as the Kirchhoff's Current Law(KCL), the junction rule states that at every junction the sum of the currents that enter is equivalent to the sum of the currents that leave. This rule also holds for a point in a line instead of a junction of several lines. We can understand this as when the flow of current is steady, the charges do not accumulate at any junction or point in a line. Thus, the total current(rate of flow of charge) flowing in has to be the same as the total current flowing out.

• Loop Rule

Also known as the Kirchhoff's Voltage Law (KVL), the loop rule states that the algebraic sum of the potential changes around any closed loop, with cells and resistors is zero. This rule holds because the electric potential depends on the location of a point. Starting from any point, if we return to the same point, the total change must be zero. 

Kirchhoff's first law is also known as Kirchhoff's junction rule. The junction rule states that the total sum of the current in a junction is equal to the total sum of currents outside it.

In simple terms, it means that the total current entering a junction is equal to the total current leaving it. There is no energy loss at the intersection.

This property of zero energy loss in Kirchoff's law is known as Conservation of charge.

Thus, I(enter) + I(exit) = 0

In the above figure, we can see that the I1, I2, and I3 are the currents entering the junction and are positive. The currents I4 and I5 are leaving the junction. Hence, they are negative. 

Hence, I₁ + I₂ + I₃ - I_{4 -} I₅ = 0

The advantage of using Kirchhoff's first law is that it can also be used in parallel circuits.

Kirchhoff's second law is also known as Kirchhoff's voltage law. It states that the sum of the total voltage in a closed network is always equal to 0.

This property of Kirchhoff's second law is also known as the conservation of energy.

In the above figure, we can see that the voltage changes either to positive or negative and returns at a single point. The direction needs to be taken either clockwise or anti-clockwise. A change in direction will not lead to the final voltage being equal to zero. 

The advantage of using Kirchhoff's second law is that it can also be applied for circuits in a series.

However, it does not take into account the fluctuations caused by the magnetic fields. Varying magnetic fields can influence the electric fields, and Kirchoff's laws may not apply during such a case.

Kirchhoff's Law in Class 12

In class 12, the chapter current electricity states and explains Kirchhoff's current and Kirchhoff's voltage law. It provides a detailed explanation of these laws and includes various applications of these laws.

Current electricity has a weightage of 7-8 marks in the exam. Hence, two or three questions from this chapter may be asked.

1. Find the current i3, as shown below

Solution: Currents i1 and i2 are flowing towards the node, and the currents i3 and i4 are flowing out of the node.

Apply Kirchhoff's law of current:

i1+i2 = i3+i4

Substituting,

2+9=i3+4

i3=7 A

2. Calculate the currents i3 and i4, as shown below.

Solution:  i1 flows into N1, and i2 and i3 flow out of N1, hence,

i1 = i2 + i3

Substituting

5 = 9 + i3

i3=−4

Since i3 is negative, it flows into node N1

At node N2,

i3+i5 = i4

Substituting

−4+10 = i4

i4=6

3. See the image below. 

Solution: Applying Kirchhoff’s second law to loop 1,

e1−Vr1−VR2=0

Substituting,

20−5−VR2=0

VR2=15 A

Apply Kirchhoff's second law to loop 2,

Vr2+e2−VR3=0

Substituting,

15+10−VR3=0

VR3=25

4. Given below is an electrical circuit. If I₁ is 1A, find the value of I₂ and I₃. 

 

Solution: From the circuit, at 'a', the departing current is I₁+I₂ while the current that enters is I₃.

The junction rule says I₃=I₁+I2. 

I₁ is the entering current at point ‘h’ and only one current leaves h.

By the junction rule, it is I₁. 

For the loop 'ahdcba', the loop rule gives 

-30I₁-41I₃ +45=0

Since I₁=1A, thus I₃=15/41=0.36A

For the loop 'ahdefgba', the loop rule gives 

-30I₁+21I₂-80=0

Since I₁=1A, thus I₂=110/21=5.23A.

5: A 10V battery connected in a cubical network consists of 12 1Ω resistors. Determine equivalent resistance and the current by each edge of the cube.

Solution: AA', AD and AB are symmetric hence the current in each is equal. Taking a closed-loop ABCC'EA, we apply the loop rule:

-IR-(1/2)IR-IR+ε=0

ε=(5/2)IR

R_eq=ε/3I=(5/6)R

R=1Ω, therefore R_eq=(5/6)Ω

For ε=10V, 3I=10/(5/6)=12A

Thus, I=4A.

The current in each edge is determined from the diagram.

6 Find the currents I₁, I₂ and I₃ from the electrical network given below.

Solution: First we apply KCL at every junction to mark the unknown current. Then we apply KVL to determine the values of I1, I2, I3.

For loop ADCA, 

10-4(I₁-I₂)+2(I₂+I₃-I₁)-I1=0

i.e, 7I₁-6I₂-2I₃=10

For loop ABCA,

10-4I₂-2(I₂+I₃)-I₁=0

i.e, I₁+6I₂+2I₃=10

For loop BCDEB,

5-2(I₂+I₃)-2(I,+I₃-I₁)=0

i.e, 2I₁-4I₂-4I₃=-5

From these three equations we get, 

I ₁ =2.5A, I ₂ =5/8A, I ₃ =15/8A.

[Image courtesy: NCERT]