Question

This question relates to the electric circuit above.
Through which point does half the total current pass?

& \left( A \right)A \\\ & \left( B \right)B \\\ & \left( C \right)C \\\ & \left( D \right)D \\\ & \left( E \right)E \\\ \end{aligned}$$

Answer 1

Hint : Here in this question we are going to use series and parallel resistance formula .There is a relationship between resistance, voltage and current.First, we will find equivalent resistance by adding when in series, an in parallel .After that we will use kirchoff’s law to find through which point half the total current passes.

Complete step-by-step solution:
A circuit is a visual display of current passing through wires.There are three basic types of electrical circuit such as series, parallel,series/parallel.A circuit is a complete path around which electricity flows. Battery is a must as source of electricity and the material of path of circuit must be of conductor type through which current should pass. To function a simple circuit necessary components are source of voltage,conductive path and resistor. There are Two forms of electrical power used by circuit are alternating current and direct current.Every circuit has a circuit breaker which works when there is imbalanced current flow or any part of component fails so,this failure causes damage to other components that is why circuit breaker is wired into a circuit called fuse.There is a source where electrons enter ,and the place where electron exits are called return or grounded earth.
Circuits in houses are generally wired in parallel. Resistance is the hindrance in the flow of current in the conductor.It opposes the flow of electrons.It is measured in ohms.
We have to calculate the equivalent resistance between point B and D in which $6\Omega \And 6\Omega$are in series and $12\Omega$in parallel. So first solve $6\Omega \And 6\Omega$in series we get,
${{R}_{2}}=6+6=12\Omega$
The resultant of $6\Omega \And 6\Omega$ is in parallel with$12\Omega$,we get
$\dfrac{1}{{{R}^{1}}}=\dfrac{1}{12}+\dfrac{1}{12}=6\Omega$
Now we calculate equivalent resistance between point B and E we get,
$R=6+{{R}^{1}}=6+6=12\Omega$
Now we are going to use kirchhoff's law for closed mesh ABDEA which is:
$IR=6+6$
$I=\dfrac{12}{R}=\dfrac{12}{12}=1A$
The above equation is the total current of the circuit.
Now we have to find voltage drop between B and D that is:
${{V}_{BD}}=I{{R}^{1}}=1\times 6=6V$
Current in the resistance is $12\Omega$
${{I}^{1}}=\dfrac{{{V}_{BD}}}{12}=\dfrac{6}{12}=0.5A$
Here, current which passes through point C in the resistance of$6\Omega$, which are in series current will be same in both resistances:
${{I}_{2}}=\dfrac{{{V}_{BD}}}{6}+6=\dfrac{6}{12}=0.5A$
Hence from the above equation it is clear that through point C half of the total current passes.
So, the correct option is C.

Note: Electricity is also found in nature in the form of lightning which occurs when large amounts of electrical energy are formed in the clouds through wind, when it occurs a large flash of electricity can be seen in the sky. Electricity travels at the speed of light. It is also a part of magnetism.