Kirchhoff’s Laws

Kirchhoff’s laws are two fundamental principles in electricity that describe the behavior of currents and voltages in electrical circuits.

• Kirchhoff’s First Law (Junction Law)

The algebraic sum of the currents entering a junction is zero.

• Kirchhoff’s Second Law (Loop Law)

The algebraic sum of the voltages around a closed loop is zero.

These formulas provide a solid foundation for understanding and solving problems related to electricity in JEE Main. By mastering these concepts, students can effectively analyze electrical circuits, calculate currents and voltages, and design electrical systems.

JEE Main Work, Energy and Power Solved Examples 

1. A particle travels in a straight path, with retardation proportional to displacement. Calculate the kinetic energy loss for every displacement x�.

Sol:

It is given that retardation(a�) is proportional to displacement (x�). So in order to solve this problem, we first write retardation in terms of velocity and after separating and integrating the quantities in the given relation, we will be able to reach the answer.

According to the question,

−a∝x−�∝�

−a=kx−�=��………..(1)

Here, k� is the proportionality constant.

Now putting a=dvdt�=d�d� in the eq.(1), we get:

−a=−dvdt=kx−�=−d�d�=��

Now after multiplying and dividing on L.H.S by dxd�, we get:

dvdtdxdx=−kxd�d�d�d�=−��

As dxdt=vd�d�=�, therefore, vdv=−kxdx�d�=−��d�

∫0vvdv=−k∫0xxdx∫�0���=−�∫�0���

[v22]vu=−k[x22]x0[�22]��=−�[�22]0�

12mv2−12mu2=−kx2m212��2−12��2=−��2�2

Thus, the loss in kinetic energy is ΔK=−kmx22Δ�=−���22.

Trick: Use the given condition of the question and put the retardation formula in terms of velocity then separate the similar quantities and integrate them.

2. A body falls to the earth from a height of 8,m8,� and then bounces to a height of 2,m2,�. Calculate the ratio of the body’s velocities right before and after the contact. Calculate the body’s kinetic energy loss as a percentage during the contact with the ground.

Sol:

It is given that heights h1=8mℎ1=8� and h2=2mℎ2=2�. Let v1�1 be the velocity of the body just before collision with the ground and v2�2 be the velocity of the body just after collision.

Now, according to the conservation of mechanical energy, we can write:

12mv21=mgh112��12=��ℎ1 and 12mv22=mgh212��22=��ℎ2

After dividing the above equation, we get:

v21v22=h1h2�12�22=ℎ1ℎ2

Putting the values of h1ℎ1 and h2ℎ2, we get:

v21v22=82=4�12�22=82=4

v1v2=2�1�2=2

Therefore, the ratio of the velocities is 2:1.

Now, the percentage loss in kinetic energy is given as:

% age loss in K.E=(K1−K2K1×100% age loss in K.E=(�1−�2�1×100 

% age loss in K.E=mg(h1−h2)mgh1×100% age loss in K.E=��(ℎ1−ℎ2)��ℎ1×100

% age loss in K.E=(8−2)8×100% age loss in K.E=(8−2)8×100

% age loss in K.E=75%% age loss in K.E=75%

Hence, the percentage loss in kinetic energy is 75 %.

Key point: The laws of conservation of mechanical energy is an important concept and so is the loss of kinetic energy expression.