Question

Derive the derivation of equivalent resistance when three resistors are connected in Parallel. (Draw Diagram) (3x1=3)

Answer 1

• We have to Derive the Derivation of equivalent resistance when three resistors are connected in Parallel .

Proof :-

• All three resistances are connected parallel to one another between the same two points .

$\bullet\sf\ Let\ the \ resistance\ be \\ \sf\ \ R_1\ ,R_2\ and\ R_3$

• Voltage remains constant

$\bullet\sf\ \ Voltage\ Across\ all\ three\ resistances\\ \\ \sf\ \ \Big\lgroup V_1=V_2=V_3\Big\rgroup$

• Current gets distributed

$\bullet\sf\ Let\ the \ current\ be \\ \sf\ \ I_1\ ,I_2\ and\ I_3\\ \\ \sf\ \ Total\ Current\ I= I_1+I_2+I_3$

• Now , By applying Ohm's Law

$\underline{\boxed{\sf\ \ V=IR}}\\ \\ \sf\ \ \ \ I= \dfrac{V}{R}\ \ \ ---eq.(i)$

$\dashrightarrow\sf\ V_1=I_1\times R_1\\ \\ \\ \dashrightarrow\sf\ I_1=\dfrac{V_1}{R_1}\ \ ----eq.(ii)\\ \\ \\ \dashrightarrow\sf I_2=\dfrac{V_2}{R_2}\ \ ----eq.(iii)\\ \\ \\\dashrightarrow\sf I_3=\dfrac{V_3}{R_3}\ \ ----eq.(iv)$

$\bullet\sf\ V_{(total)}=I_{(total)}\times R_{(total)}\\ \\ \sf\ \ I_{(total)}=\dfrac{V_{(total)}}{R_{(total)}}\\ \\ \\ \dashrightarrow\sf\ \dfrac{V_t}{R_t}=I_t\\ \\ \\ \dashrightarrow\sf\ \dfrac{V_t}{R_t}= \dfrac{V_1}{R_1}+\dfrac{V_2}{R_2}+\dfrac{V_3}{R_3}\ \ \ \ \ \Big[I_t=I_1+I_2+I_3\Big]\\ \\ \\ \dashrightarrow\sf\ \dfrac{V_t}{R_t}=\dfrac{V_t}{R_1}+\dfrac{V_t}{R_2}+\dfrac{V_t}{R_3}\ \ \ \ \Big[V_t=V_1=V_2=V_3\Big]\\\ \\ \\ \dashrightarrow\sf\ \dfrac{\cancel{V_t}}{R_t}=\cancel{V_t}\bigg\lgroup \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\bigg\rgroup\\ \\ \\ \dashrightarrow\sf\ \dfrac{1}{R_t}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$

• Thus , If three resistance are connected in parallel,then their resultant resistance R is

$\underline{\boxed{\sf\ \ \dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}}$