Question

The value of current $'i'$ for the given circuit is

A. $10{\text{ A}}$
B. ${\text{5 A}}$
C. ${\text{2}}{\text{.5 A}}$
D. $20{\text{ A}}$

Answer 1

Generally in order to find out the effective resistance we use series and parallel connection formulas. In case there is a symmetry then we use mirror symmetry or perpendicular axes symmetry and we can use star delta conversion in some cases to find out the effective resistance.
Formula used:
\eqalign{ & {R_s} = {R_1} + {R_2} + {R_3} + ......... \cr & \dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ....... \cr & V = i{R_{eff}} \cr}

Complete answer:
Rate of flow of charge is known as current. A material which allows current to pass through it is known as a conductor. No conductor will be perfect. It will have some resistance. The property to hinder the flow of current is called resistance and a device which does that is known as a resistor.
If the same current is passing through all resistors then we tell those are connected in series. If potential difference is the same for all resistors then those resistors are told to be in parallel.

The given circuit can be solved by using wheatstone bridge principle. If a, b, c, d are the resistances in the given diagram then
If $\dfrac{a}{b} = \dfrac{c}{d}$ then potential at terminal A will be equal to potential at point B.
Current passing through the resistor is given as $V = i{R_{eff}}$. Where V is the potential difference. Since
${V_A} = {V_B}$ potential difference will be zero. Hence current through branch AB will be zero. Then the effective circuit will be below one.

Since wheatstone bridge condition is satisfied we removed the central resistance.
When resistors are connected in series then effective resistance is ${R_s} = {R_1} + {R_2} + {R_3} + .........$
When resistors are connected in parallel effective resistance will be $\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .......$
Top most branch resistors of rhombus are in series connection. Hence effective resistance is $5 + 5 = 10$
bottom most branch resistors of rhombus are also in series connection. Hence effective resistance is $5 + 5 = 10$
Now two branches will be in parallel so effective resistance is
\eqalign{ & \dfrac{1}{R} = \dfrac{1}{{10}} + \dfrac{1}{{10}} \cr & \Rightarrow R = \dfrac{{10}}{2} \cr & \Rightarrow R = 5 \cr}
Now this resistance will be in parallel again with bottom most remaining resistor so total effective resistance will be
\eqalign{ & \dfrac{1}{{{R_{eff}}}} = \dfrac{1}{5} + \dfrac{1}{5} \cr & \Rightarrow {R_{eff}} = \dfrac{5}{2} \cr & \Rightarrow {R_{eff}} = 2.5 \cr}
We have the equation $V = i{R_{eff}}$. V is 25volts here and ${R_{eff}} = 2.5$
Hence total current will be
$V = i{R_{eff}}$
\eqalign{ & \Rightarrow 25 = i(2.5) \cr & \Rightarrow i = 10 \cr}
When two equal resistors are connected in parallel total current will divide into half and pass through each resistor.
Total current is 10 ampere out of which half will pass through the bottom most Resistor and the remaining half will pass to that rhombus resistor connection.
So $\dfrac{{10}}{2} = 5A$ i.e 5 amperes is our answer.

Hence option B will be correct.

Note:
By using wheat stone bridge it became easy for us to solve the problem. If they ask us to find out the total current flown in the circuit then the answer would be 10 amperes. But they asked us to find out which current is passing to the rhombus shaped resistor circuit hence it is 5 amperes.