In the given circuit find the current through the \[6\Omega \]resistance.

A.10 A
B. 7 A
C. 25 A
D. 30 A

Hint:Before we proceed with the problem, it is important to know about the half-life of a radioactive substance. Kirchhoff’s Current law states that the sum of the incoming currents flowing into the circuit is equal to the sum of outgoing currents flowing out of the circuit.

Formula Used:
By Kirchhoff’s Current Law (KCL) we have
\[i = {i_1} + {i_2}\]…….. (1)
Where,
\[i\] is the current flowing through \[6\Omega \].
\[{i_1}\] is the current flowing through\[20\Omega \].
\[{i_2}\] is the current flowing through \[5\Omega \].

Complete step by step solution:
We are able to solve the problem step by step as follows.

The circuit consists of three resistors of value, \[6\Omega \], \[20\Omega \] and \[5\Omega \] with two batteries of 140 V and 90 V. in order to find the current, first we need to find the potentials at lower points or ground points. Since it is grounded the potential is zero at these points. At point A the potential is 140V volts and at point B the potential is 90V.

At the middle point assume that the potential is V. Due to these batteries the current \[{i_1}\] and \[{i_2}\] flows towards point V through the resistors \[20\Omega \] and \[5\Omega \]. Here\[i\] is the current across \[6\Omega \].

Now it is easy to find the current \[i\]using the equation (1) i.e.,
\[i = {i_1} + {i_2}\]
\[\dfrac{{V - 0}}{6} = \dfrac{{140 - V}}{{20}} + \dfrac{{90 - V}}{5} \\ \] …….. (2)
\[\Rightarrow \dfrac{V}{6} = \dfrac{{140 - V + 360 - 4V}}{{20}} \\ \]
\[ \Rightarrow \dfrac{V}{6} = \dfrac{{500 - 5V}}{{20}} \\ \]
\[ \Rightarrow \dfrac{V}{6} = \dfrac{{100 - V}}{4} \\ \]
\[ \Rightarrow 4V = 600 - 6V\]
\[ \Rightarrow V = 60V\]
Then finding the value of \[i\] using equations (1) and (2) we get,
\[i = \dfrac{{V - 0}}{6}\]
\[ \Rightarrow i = \dfrac{{60 - 0}}{6}\]
\[ \therefore i = 10A\]
Therefore, the current through the \[6\Omega \] resistance is \[10A\].

Hence, Option A is the correct answer

Note:Kirchhoff's current law is not suitable for AC circuits of high frequencies and for time-varying magnetic fields. KCL is applied only when the electric charge in a circuit is constant.