Question

A 12v battery is connected to a circuit having 3 resister of resistance 2ohm,3ohm,6ohm in parallel. calculate the total resistance of a circuit. find the current passing through each resister
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Answer 1

Given :

• A 12v battery is connected to a circuit having 3 resistor of resistance 2Ω, 3Ω and 6Ω in parallel.

To Find :

• The total resistance of a circuit.
• The current passing through each resistor.

According to the question,

As we know that 3 resistor of resistance 2Ω, 3Ω and 6Ω are connected in parallel. So we will use the formula,

$\longrightarrow\sf{ \dfrac{1}{R} = \dfrac{1}{R _{1} } + \dfrac{1}{R _{2}} + \dfrac{1}{R _{3}} }$

$\longrightarrow\sf{ \dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} }$

$\longrightarrow\sf{ \dfrac{1}{R} = \dfrac{3 + 2 + 1}{6} }$

$\longrightarrow\sf{ \dfrac{1}{R} = \dfrac{6}{6}}$

$\longrightarrow \sf{R\: = 1}$

• So,the total resistance of a circuit is 1 Ω.

Now,

• V = IR

Where,

• V = Voltage
• I = Current
• R = Resistance

➝ V = IR

➝ 12 = I × 2

➝ 6 = I

➝ I = 6 Ampere

➝ V = IR

➝ 12 = I × 3

➝ 4 = I

➝ I = 4 Ampere

➝ V = IR

➝ 12 = I × 6

➝ 2 = I

➝ I = 2 Ampere.

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Answer 2

Given that , A 12 V ( Volt ) battery is connected to a circuit , The resistor of Resistance 2 Ω , 3 Ω & 6Ω are connected in parallel.

Exigency To Find : The Total ( or Equivalent ) Resistance of a circuit & the current passing through each resistor ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀$\underline{ \purple{\bf \:\:\bigstar \: Finding \:Total\:( \:or \:Equivalent \:) \: \:Resistance \::\:}}$

As , We know that ,

⠀⠀⠀⠀⠀When , The resistors are connected in parallel, then Total ( or Equivalent ) Resistance of a circuit will be ,

$\qquad \dag\:\:\bigg\lgroup \pmb{\bf{ \:\:\dfrac{1}{R_{eq}} \:=\: \dfrac{1}{R_1 } + \:\dfrac{1}{R_2} \:+ \:\dfrac{1}{R_3} \: }}\bigg\rgroup$

⠀⠀⠀⠀⠀Here $\bf R_{eq}$ is the Total Resistance, $\bf \:R_1 \:,\: R_2 \:\:\& \: R_3 \:$ are the resistors which are connected in parallel .

$\qquad \dashrightarrow \sf \dfrac{1}{R_{eq}} \: =\: \dfrac{1}{R_1} \:+ \:\dfrac{1}{R_2 } \:+ \:\dfrac{1}{R_3} \:$

⠀⠀⠀⠀⠀The resistor of Resistance 2 Ω , 3 Ω & 6Ω are connected in parallel.

$\qquad \dashrightarrow \sf \dfrac{1}{R_{eq}} \: =\: \dfrac{1}{R_1} \:+ \:\dfrac{1}{R_2 } \:+ \:\dfrac{1}{R_3} \:\\\\\qquad \dashrightarrow \sf \dfrac{1}{R_{eq}} \: =\: \dfrac{1}{2} \:+ \:\dfrac{1}{3} \:+ \:\dfrac{1}{6} \:\\\\\qquad \dashrightarrow \sf \dfrac{1}{R_{eq}} \: =\: \dfrac{3 \:+\:2\:+\:+\:1}{6} \:\\\\\qquad \dashrightarrow \sf \dfrac{1}{R_{eq}} \: =\: \dfrac{6}{6} \:\\\\\qquad \dashrightarrow \sf \dfrac{1}{R_{eq}} \: =\: \cancel {\dfrac{6}{6}} \:\\\\\qquad \dashrightarrow \sf R_{eq} \: =\: 1 \:\\\\\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ \:R_{eq} \: =\: 1 \:}}}}}\:\:\bigstar \:$

$\qquad \therefore \:\underline {\sf \: Hence, \:The \:Total \:Resistance \: will \; be \: \pmb{\bf 1 \: \Omega \:}\:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀$\underline{ \purple{\bf \:\:\bigstar \: Finding \:Amount \:of \:Current \:through \:each \:Resistor \::\:}}$

As, We know that ,

• Formula for CURRENT :

$\qquad \dag\:\:\bigg\lgroup \pmb{\bf{ \:\:V \: = \:I \: R \: }}\bigg\rgroup$

⠀⠀⠀⠀⠀Here , V is the voltage of Battery , I is the amount of Current & R is the Resistance .

$\qquad \underline {\purple {\maltese \:\bf 2 \:\Omega \:}}\\\\\qquad \dashrightarrow \sf \:\:V \: = \:I \: R \:$

• The V or Voltage of Battery is 12 V &
• The R or Resistance of Resistor is 2 Ω .

$\qquad \dashrightarrow \sf \:\:V \: = \:I \: R \: \\\\\qquad \dashrightarrow \sf \:\:12 \: = \:I \:(2) \: \\\\\qquad \dashrightarrow \sf \:\:I \: = \:\dfrac{12}{2} \: \\\\\qquad \dashrightarrow \sf \:\:I \: = \:6\: \\\\\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ \:I \:( \:or \: Current \:) \: =\: 6 \:Ampere \:}}}}}\:\:\bigstar \:$

• The Current passing through 2 Ω is 6 Ampere.

⠀⠀⠀⠀⠀

$\qquad \underline {\purple {\maltese \:\bf 3 \:\Omega \:}}\\\\\qquad \dashrightarrow \sf \:\:V \: = \:I \: R \:$

• The V or Voltage of Battery is 12 V &
• The R or Resistance of Resistor is 3 Ω .

$\qquad \dashrightarrow \sf \:\:V \: = \:I \: R \: \\\\\qquad \dashrightarrow \sf \:\:12 \: = \:I \:(3) \: \\\\\qquad \dashrightarrow \sf \:\:I \: = \:\dfrac{12}{3} \: \\\\\qquad \dashrightarrow \sf \:\:I \: = \:4\: \\\\\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ \:I \:( \:or \: Current \:) \: =\: 4 \:Ampere \:}}}}}\:\:\bigstar \:$

• The Current passing through 3 Ω is 4 Ampere.

⠀⠀⠀⠀⠀

$\qquad \underline {\purple {\maltese \:\bf 6 \:\Omega \:}}\\\\\qquad \dashrightarrow \sf \:\:V \: = \:I \: R \:$

• The V or Voltage of Battery is 12 V &
• The R or Resistance of Resistor is 2 Ω .

$\qquad \dashrightarrow \sf \:\:V \: = \:I \: R \: \\\\\qquad \dashrightarrow \sf \:\:12 \: = \:I \:(6) \: \\\\\qquad \dashrightarrow \sf \:\:I \: = \:\dfrac{12}{6} \: \\\\\qquad \dashrightarrow \sf \:\:I \: = \:2\: \\\\\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\pink{ \:I \:( \:or \: Current \:) \: =\: 2 \:Ampere \:}}}}}\:\:\bigstar \:$

• The Current passing through 6 Ω is 2 Ampere.

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$\qquad\quad\underline{\bigstar\:{\pmb{\mathbb{ADDITIONAL\:INFORMATION \:\:: \: }} }}$

▪︎⠀To Calculate Current ( I ) , we use ;

$\qquad \qquad \leadsto \sf I \:=\:\dfrac{ \:Q \:}{\:T\:}\:\:$

⠀⠀⠀⠀⠀Here I is the Current, Q is the Electrical charge & T is the Total Time taken .

▪︎⠀To Calculate Voltage ( V ) of Battery , we use ;

$\qquad \qquad \leadsto \sf V \:=\:\dfrac{ \:E \:}{\:Q\:}\:\:$

⠀⠀⠀⠀⠀Here , V is Voltage of Battery, E is the Energy & W is the amount of Work done .

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