Question

2 ohm resistor and 6 ohm resistor are connected in series with a 4 V battery. Calculate amount of charge passing through the battery in 10 seconds​

Answer 1

Answer:

5 C if current is passing through the circuit in 10 seconds.

Explanation:

Resistors = 2 Ω and 6 Ω

Battery of = 4 Volts

Time = 10 seconds

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★ Equivalent Resistance =

$\sf{\longrightarrow} \: R_s =R_1 + R_2$

• $\sf{R_1}$ = 2 Ω
• $\sf{R_2}$ = 6 Ω

$\sf{\longrightarrow} \: R_s = 2 + 6$

$\sf{\longrightarrow} \: R_s = 8 \:ohms$

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★ Current in the circuit =

$\sf{\longrightarrow} \: I = \dfrac{V}{R}$

• V = 4 V
• R = 8 Ω

$\sf{\longrightarrow} \: I = \dfrac{4}{8}$

$\sf{\longrightarrow} \: I = \dfrac{1}{2} \: A$

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★ Amount of charge passing:

$\sf{\longrightarrow} \: Q=IT$

• I = 1/2 A
• T = 10 seconds

$\sf{\longrightarrow} \: Q= \dfrac{1}{2} \times 10$

$\sf{\longrightarrow} \: Q=5 \: coulomb$

Therefore, 5 C if current is passing through the circuit in 10 seconds.

Answer 2

Given : 2 ohm resistor and 6 ohm resistor are connected in series with a 4 V ( or Volt ) 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 [ R ].

⠀⠀⠀⠀⠀Therefore,

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

As , We know that ,

$\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:R_1 \:(Resistance\:1 )\:+ R_2\:(Resistance \:2 )\:\:$

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

$\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:R_1 + R_2\:\:\:$

$\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:2 + 6 \:\:\:$

$\qquad:\implies \sf R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:8 \:\Omega \:\:\:$

$\qquad \therefore \pmb{\underline{\purple{\:R_eq\:\:( \:Equivalent \:Resistance \:)\: \:=\:8 \:\Omega }} }\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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

As , We know that ,

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

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

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

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

$\qquad \dashrightarrow \sf Current \:(I)\:\:= \:\dfrac{4\:}{8\:}$

$\qquad \dashrightarrow \sf Current \:(I)\:\:= \:\dfrac{1\:}{2\:}$

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

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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

As , We know that ,

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

$\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:I( Current)\:\times \:T(Time)\:\:$

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

$\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:I( Current)\:\times \:T(Time)\:\:$

$\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:\dfrac{1}{2} \:\times \:10\:\:$

$\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:\dfrac{10}{2} \:\:\:$

$\qquad \dashrightarrow \sf Charge \:(Q)\:\:= \:5\:\:\:$

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

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