Question

Calculate I for a given circuit.

A. $10A$
B. $5A$
C. $2.5A$
D. $20A$

Answer 1

Hint: According to the ohms law the current flowing through a circuit is directly proportional to the voltage drop across the circuit. This proportionality is resolved by a proportionality constant R which represents the resistance of the circuit. The resistance is a linear property which remains constant in constant external conditions and depends on the internal factors such as area and the length of the material.

Complete step by step answer:
Wheatstone bridge is a circuit connection where 4 resistances are in rhombus type shape as shown in the circuit below,
 

The balance condition is when the ratio of resistance 1 and 2 is equal to the ratio of resistances 3 and 4 mathematically,
$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{R}_{3}}}{{{R}_{4}}}$
According to the principle of Wheatstone bridge, under balance conditions no current passes through the terminal C and D.
When we see the circuit given in the question we can conclude that one of the sections of the circuit is a balanced condition of Wheatstone bridge.
$\dfrac{5}{5}=\dfrac{5}{5}=1$
So, no current passes through the residence between terminals C and D.
The above circuit can be simplified as,

So the total resistance between points A and B is
$\frac{1}{R}=\frac{1}{{{R}_{1}}+{{R}_{3}}}+\frac{1}{{{R}_{2}}+{{R}_{4}}}$
This is because resistance in series is given by the algebraic sum of the resistance in series. Mathematically,
$R={{r}_{1}}+{{r}_{2}}+...$
So the resistance between the points A and B will be,
$\begin{align}
 & \dfrac{1}{R}=\dfrac{1}{5+5}+\dfrac{1}{5+5} \\
 & \Rightarrow R=\frac{10}{2} \\
 & \Rightarrow R=5\Omega \\
\end{align}$
Terminal A and B are parallel to the resistance of \[5\Omega \]. So the total resistance of the circuit is given by
\[\begin{align}
 & \frac{1}{{{R}_{total}}}=\dfrac{1}{5}+\dfrac{1}{5} \\
 & \Rightarrow {{R}_{total}}=\dfrac{5}{2} \\
 & \Rightarrow {{R}_{total}}=2.5\Omega \\
\end{align}\]
So the total resistance of the circuit is $2.5\Omega $
By using the application of the ohm's law we know that
$V=IR$
So, by rearranging the above equation,
$\begin{align}
 & I=\dfrac{V}{R} \\
 & \Rightarrow I=\dfrac{25}{2.5} \\
 & \therefore I=10A \\
\end{align}$
So the total current flowing through the circuit is $10A$.

So, the correct answer is “Option A”.

Note: The Wheatstone bridge is named after the person who discovered it, Mr. Charles Wheatstone. It was formed to measure the value of some unknown resistance value and a means of calibrating measuring instruments such as voltmeters, ammeters, etc., by the use of long resistive slide wire. Wheatstone bridge is still used to measure the low value of resistance.