Question

Kirchhoff 1st and 2nd laws are based on conservation of
A.) Energy and charge respectively
B.) Charge and energy respectively
C.) Mass and charge respectively
D.) None of the above

Answer 1

Hint: Kirchhoff’s first law is given by $\sum\limits_{}^{} {I = 0} $ , here I is the current and Kirchhoff’s second law is given by $\sum\limits_{}^{} {IR = 0} $ , where I and R are the current and resistance.

Complete step-by-step answer:

Formula used - $\sum\limits_{}^{} {I = 0} $ , $\sum\limits_{}^{} {IR = 0} $ , $I = \dfrac{Q}{t}$
Let us first discuss what are Kirchhoff's 1st and 2nd laws.

Kirchhoff’s first law-

It states that the sum of current entering to the junction is equal to sum of current leaving the junction i.e., $\sum\limits_{}^{} {I = 0} $ , where I denote the current.

Kirchhoff’s second law-

It states that the algebraic sum of potential drops in a closed circuit is zero, i.e., $\sum\limits_{}^{} {IR = 0} $ , where I and R are the current and the resistance.

Now, as we have seen what are the two laws about. So, now we can see that-

The first law of Kirchhoff’s is based on charge conservation, as it talks about the summation of current to be zero at any junction, which means that if current is conserved that implies that charge is also conserved. We can say this because we know, current is the charge per unit time, so if current is conserved then charge is conserved.

Since, $\sum\limits_{}^{} {I = 0} $

so, using $I = \dfrac{Q}{t}$ , we can write-

$
  \sum\limits_{}^{} {\dfrac{Q}{t} = 0} \\
  \sum\limits_{}^{} {Q = 0} \\
$

So, the first law is based on charge conservation.

Now, if we see about Kirchhoff’s second law-

The second law is based on conservation of energy, as the sum of potential drops in a closed circuit is zero, i.e., $\sum\limits_{}^{} {IR = 0} $ , we can also write
 $\sum\limits_{}^{} {V = 0} $ , as V = IR.

Now, we can say that energy is conserved because a circuit loop is a closed conducting path so no energy is lost. We can also say that as moving around a closed loop, or circuit, you will end up at the same point where you started and therefore, we can say back to the same potential with no loss of voltage around the loop and hence no loss of energy and energy is conserved.

Hence, the Kirchhoff’s 1st and 2nd law are based on charge and energy conservation.

So, the correct option is B.

Note: Whenever such types of questions appear, then always first understand what the Kirchhoff’s first and second law states and then using the results see whether there is conservation of charge or energy. As mentioned in the solution, the first law is based on charge conservation and second on energy conservation.