Question

2 ohm resistor and a 6 ohm resistor connected in series with

a 4 v battery. calculate the amount of charge passing through the battery in 10

seconds.​

Answer 1

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2 ohm resistor and a 6 ohm resistor connected in series with a 4 v battery calculate the amount of charge passing through the battery in 10 seconds.

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: 2 ohm resistor and 6 ohm resistor are connected in series with a 4 V ( orVolt) battery and the Total time taken is 10 s ( or Second ) .

Exigency To Find : The amount of charge passing through the battery .

⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀⠀The battery of is of 4 V ( or Volt ) [ V ] .

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀ The Total Time taken [ T ] is 10 s ( or Seconds ) .

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀2 ohm resistor and 6 ohm resistor are connected in series

⠀Therefore,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀EQUIVALENT RESISTANCE will be :

As , We know that ,

$\begin{gathered}\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:R_1 \:(Resistance\:1 )\:+ R_2\:(Resistance \:2 )\:\:\\\\\end{gathered} :⟹ q(EquivalentResistance)=R 1 (Resistance1)+R 2 (Resistance2)$

$\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered} ⋆NowBySubstitutingtheknownValues:$

$\begin{gathered}\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:R_1 + R_2\:\:\:\\\\\end{gathered} :⟹R e q(EquivalentResistance)=R 1 +R 2$

$\begin{gathered}\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:2 + 6 \:\:\:\\\\\end{gathered} :⟹R e q(EquivalentResistance)=2+6$

$\begin{gathered}\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:8 \:\text{\O}mega \:\:\:\\\\\end{gathered} :⟹R e$

q(EquivalentResistance)=8Ømega

$\begin{gathered}\qquad \therefore \pmb{\underline{\purple{\:R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:8 \:\text{\O}mega }} }\:\:\bigstar \\\end{gathered} ∴R e$

★

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Firstly, We need to find the Amount of Current in circuit :

As , We know that ,

$\begin{gathered}\qquad \dag \:\:\: \boxed {\underline {\pink{ \pmb { Current \:(I)\:\:= \:\dfrac{Volt \:\:(V)\:}{Resistance \:(R)\:} }}}}\\\\\end{gathered} †$

$\begin{gathered}\qquad \dashrightarrow \sf Current \:(I)\:\:= \:\dfrac{ Volt \:(V) \:}{Resistance \:\:(R)\:} \\\\\end{gathered} ⇢Current(I)= Resistance(R)Volt(V)$

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

$\begin{gathered}\qquad \dashrightarrow \sf Current \:(I)\:\:= \:\dfrac{Volt \:\:(V)\:}{Resistance \:(R)\:} \\\\\end{gathered} ⇢Current(I)= Resistance(R)Volt(V)$

$\begin{gathered}\qquad \dashrightarrow \sf Current \:(I)\:\:= \:\dfrac{4\:}{8\:} \\\\\end{gathered} ⇢Current(I)= 84$

$\begin{gathered}\qquad \dashrightarrow \sf Current \:(I)\:\:= \:\dfrac{1\:}{2\:} \\\\\end{gathered} ⇢Current(I)= 21$

$\begin{gathered}\qquad \therefore \pmb{\underline{\purple{\:Current \:(I)\:\:= \:\dfrac{1\:}{2\:} \:A }} }\:\:\bigstar \\\end{gathered} ∴Current(I)= 21 A$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Now , Finding the amount of Charge passing through it :

As , We know that ,

$\begin{gathered}\qquad \dag \:\:\: \boxed{ \underline {\pink{ \pmb {Charge \:(Q)\:\:= \:I( Current)\:\times \:T(Time)\: }}}}\\\\\end{gathered} †$

$\begin{gathered}\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:I( Current)\:\times \:T(Time)\:\: \\\\\end{gathered} ⇢Charge(Q)=I(Current)×T(Time)$

$\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:I( Current)\:\times \:T(Time)\:\: \\\\\end{gathered} ⇢Charge(Q)=I(Current)×T(Time)$

$\begin{gathered}\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:\dfrac{1}{2} \:\times \:10\:\: \\\\\end{gathered} ⇢Charge(Q)= 21 ×10$

$\begin{gathered}\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:\dfrac{10}{2} \:\:\: \\\\\end{gathered} ⇢Charge(Q)= 210$

$\begin{gathered}\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:5\:\:\: \\\\\end{gathered} ⇢Charge(Q)=5$

$\begin{gathered}\qquad \therefore \pmb{\underline{\purple{\:Charge \:(Q)\:\:= \:5\:\:C\:( \:or\:coulomb\:) }} }\:\:\bigstar \\\end{gathered}$

$★ \therefore \underline { \sf Hence , \: The \:amount\:of\:charge \:passing \:through \:battery\:in \: 10 \:seconds \: is \: \bf 5 \: coulomb \:}∴$

Hence,The amount of charge passing through battery in 10 second sis 5  coulomb.