Question

give the circuit diagram and calculate the current through two resistors 2ohm and 10ohm,if they are in parallel and connected to a potential difference of 6volt?
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Answer 1

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• Two 2Ω resistor and a 10Ω are connected in parallel combination .
• The potential difference is 6 volts .

$\red{\bigstar}</p><p>\underline{\underline{\textsf{\textbf{ To Find :- }}}}$

• The Current through the circuit .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

We must firstly find out the net resistance of the circuit. We know that when two resistors are connected in parallel , say R₁ and R ₂ , then the net resistance is R₁ R₂/ R₁ + R₂ . Therefore ,

$\sf\dashrightarrow R_{net}= \dfrac{ R_1R_2}{R_1+R_2} \\\\\\\sf\dashrightarrow R_{net}= \dfrac{ 2\Omega * 10\Omega }{(2+10)\Omega} \\\\\\\sf\dashrightarrow R_{net}= \dfrac{ 20\Omega^2}{12\Omega } \\\\\\\sf\dashrightarrow \boxed{\pink{\sf R_{net}= ^{5}\!/_3 \ Omega }}$

$\rule{200}3$

❒ Now using Ohm's Law :-

$\sf\dashrightarrow Voltage = Current * Resistance \\\\\\\sf\dashrightarrow 6V = I \times \dfrac{5}{3} \Omega \\\\\\\sf\dashrightarrow I = 6 * \dfrac{ 3}{5} \\\\\\\sf\dashrightarrow \underset{\blue{\sf Required\ Current }}{\underbrace{\boxed{\pink{\frak{ Current ( I ) = 3.6 \ Ampere }}}}}$

Answer 2

Given that , The Resistors of Resistance 2Ω and 10 Ω are connected in parallel combinations and there potential difference of 6 Volt .

Exigency To Find : The Current through the circuit ?

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$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Diagram :- }}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(6.5,0){2}{\line(0,1){3}}\put(0,3){\line(1,0){1}}\put(5.5,3){\line(1,0){1}}\put(0,0){\line(1,0){3}}\put(3.3,0){\line(1,0){3.2}}\put(2.7,-1.8){\bf For (I)}\put(3.2,1.5){\bf R}\put(3.2,4.4){\bf R}\put(1.5,3.9){\bf I_1 }\put(1.5,2){\bf I_2 }\put(3,-0.8){\bf\large V}\qbezier(1,3)(1,3)(2.7,4)\qbezier(5.5,3)(5.5,3)(4,4)\qbezier(1,3)(1,3)(2.7,2)\qbezier(5.5,3)(5.5,3)(4,2)\multiput(2.7,4)(0.42,0){3}{\line(1, - 2){0.2}}\multiput(2.9,3.6)(0.42,0){3}{\line(2, 5){0.19}}\multiput(2.9,2.4)(0.42,0){3}{\line(1, - 2){0.2}}\multiput(2.7,2)(0.42,0){3}{\line(2, 5){0.19}}\multiput(0,1.5)(6.3,0){2}{\line(1, - 2){0.2}}\multiput( - 0.2, 1.1)(6.7,0){2}{\line(2, 5){0.19}}\put(3, - 0.3){\line(0, 1){0.6}}\put(3.25, - 0.2){\line(0, 1){0.4}}\multiput(0.4, 1.1)(6.5,0){2}{\sf{I}}\end{picture}$

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❍ Let's Consider Resistance of two Resistors be R₁ & R ₂  , respectively.

⠀⠀⠀⠀⠀To Calculate the amount of Current passing through Circuit . Firstly, we must calculate the Total or Equivalent Resistance.

⠀⠀⠀⠀⠀¤ Calculating the Total or Equivalent Resistance of Resistor :

As We know that ,

⠀⠀⠀When two Resistors [ R₁ & R ₂ ] are connected in parallel combinations , then the Total or Equivalent Resistance will be ,

$\qquad \star \:\:\underline {\boxed {\bf{\pmb{\pink{ \:\:Total \:Resistance \:\:(\:R_{eq}\:) \:=\:\dfrac{ \: R_1 R_2 }{R_1 + R_2} }\:{\Omega \:\:\:}}}}}$

⠀⠀⠀⠀⠀Here , R₁ & R ₂ are the Resistance of two Resistors which are connected in parallel combinations.

$\qquad \dashrightarrow \sf \:R_{eq}\: \:=\:\dfrac{ \: R_1 R_2 }{R_1 + R_2} \: \Omega \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \:R_{eq}\: \:=\:\dfrac{ \: R_1 R_2 }{R_1 + R_2} \: \Omega \:\\\\\qquad \dashrightarrow \sf \:R_{eq}\: \:=\:\dfrac{ \: 2(10) }{10 + 2 } \:\Omega \:\\\\\qquad \dashrightarrow \sf \:R_{eq}\: \:=\:\dfrac{ \: 20 }{12 } \:\Omega \:\\\\\qquad \dashrightarrow \sf \:R_{eq}\: \:=\:\dfrac{ \: 5 }{3 } \:\Omega \:\\\\\qquad \dashrightarrow \sf \:R_{eq}\: \:=\:1.66 \:\Omega \:\\\\\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple {\:\:\:Total \:Resistance \:\:(\:R_{eq}\:) \: \:=\:1.66 \:\Omega }}}}}\:\:\bigstar\:$

⠀⠀⠀⠀⠀¤ Calculating the Current ( I ) Passing through Circuit :

As , We know that ,

$\qquad \star \:\:\underline {\boxed {\sf{\pmb{\pink{ \:\:Current \:\:(\:I\:) \:=\:\dfrac{ \:Potential \: Difference \:(V)}{\:\:Total \:Resistance \:\:(\:R_{eq}\:) \:}}{ \:Ampere \:(A) \: }\:\:\:}}}}$

$\qquad \dashrightarrow \sf \:\:Current \:\:(\:I\:) \:=\:\dfrac{ \:Potential \: Difference \:(V)}{\:\:Total \:Resistance \:\:(\:R_{eq}\:) \:} \:Ampere \:(A)\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \:\:Current \:\:(\:I\:) \:=\:\dfrac{ \:Potential \: Difference \:(V)}{\:\:Total \:Resistance \:\:(\:R_{eq}\:) \:} \:Ampere \:(A)\:\\\\\qquad \dashrightarrow \sf \:\:Current \:\:(\:I\:) \:=\:\dfrac{ \:6\:}{\:\:1.66 \:} \:Ampere \:(A)\:\\\\\qquad \dashrightarrow \sf \:\:Current \:\:(\:I\:) \:=\:3.61 \:Ampere \:(A)\:\\\\\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple {\:\: \:\:Current \:\:(\:I\:) \:=\:3.61 \:Ampere \:(A)\:\: }}}}}\:\:\bigstar\:$

$\qquad \therefore \underline {\:\sf\: Hence,\:The\:Current \:passing\:through \:circuit \:is\:\pmb{\bf 3.61 \:Ampere \:}\:.}$

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