Question

(a) Three resistors 2 W, 4 W and 5 W are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Answer 1

your question is ----> (a) Three resistors 2 Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

solution : (a) if resistors are connected in parallel then , equivalent resistance can be found by

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+......$

here,R₁ = 2Ω , R₂ = 4Ω and R₃ = 5Ω

so, 1/Req = 1/2Ω + 1/4Ω + 1/5Ω

1/Req = (10 + 5 + 4)/20 = 19/20

so, Req = 20/19 = 1.05Ω

hence, total resistance of the combination is 1.05Ω

(b) potential of 20V will be same across each resistor , so currrent $I_1=\frac{V}{R_1}=\frac{20}{2}=10A$

$I_2=\frac{V}{R_2}=\frac{20}{4}=5A$

$I_3=\frac{V}{R_3}=\frac{20}{5}=4A$

hence, total current drawn from the cell , $I=I_1+I_2+I_3=10A+5A+4A=19A$

Answer 2

$\huge \underline \mathrm \purple{Question↣}$

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A) Three resistors 2 W, 4 W and 5 W are combined in parallel. What is the total resistance of the combination ?

B) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.‎

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$\huge \underline \mathrm \green{Answer↣}$‎

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$\Large\rm{ A) }$ Resistors$\bold{ \: R_{1} = 1 Ω , \: R_{2} = 2 Ω \: and \: R_{3} = 3 Ω }$ are combined in parallel‎

Hence, ‎‎

The total resistance of the above circuit can be calculated by the following formula :‎‎

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⠀⠀⠀⠀⠀⠀⠀⠀${\underline{\boxed{\bold{\frac{ 1 }{ R } = \frac{ 1 }{ R_{1} } + \frac{ 1 }{ R_{2} } + \frac{ 1 }{ R_{3} }}}}}$ ‎‎‎

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⠀⠀⠀⠀⠀⠀⠀⠀⠀$\Large\sf{ : \implies \: \frac{1}{R} = \frac{1}{2} + \frac{1}{4} + \frac{1}{ 5} }$‎‎‎

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\Large\sf{ : \implies \: \frac{1}{R} = \frac{19}{20} }$

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Therefore, total resistance of the parallel combination given above is given by :

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${\sf{\Large{\red{⇝| R = \frac{ 20 }{ 19 } | }}}}$

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$\Large\rm{ B)}$Given that emf of the battery , $\bold{E = 20 V}$

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Let the current flowing through resistor $\bold{ R_{1}}$ be $\bold{ I_{1}}$

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$\bold{ I_{1}}$ is given by :

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⠀⠀⠀⠀⠀⠀⠀⠀$\large{ \boxed{ \underline{ \bold{I_{ 1} = \frac{ v}{R_{ 1}} }}}}$‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf{ : \implies \: \:I _{ 1} \: = \frac{20}{2} }$‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf\red{ : \implies \: I_{ 1} = 10A}$‎

Let the current flowing through resistor $\bold{ R_{2}}$ be $\bold{ I_{2}}$‎

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$\bold{ I_{2}}$ is given by :‎

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⠀⠀⠀⠀⠀⠀⠀⠀$\large{ \boxed{ \underline{ \bold{I_{ 2} = \frac{ v}{R_{ 2}} }}}}$ ‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf{ : \implies \: \:I _{ 2} \: = \frac{20}{4} }$‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf\red{ : \implies \: I_{ 2} = 5A}$‎

Let the current flowing through resistor $\bold{ R_{3}}$ be $\bold{ I_{3}}$‎

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$\bold{ I_{3}}$ is given by :‎

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⠀⠀⠀⠀⠀⠀⠀⠀$\large{ \boxed{ \underline{ \bold{I_{ 3} = \frac{ v}{R_{ 3}} }}}}$‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf{ : \implies \: \:I _{ 3} \: = \frac{20}{5} }$‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf\red{ : \implies \: I_{ 3} = 4A}$‎

Therefore , the total current can be found by the following formula :

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⠀⠀⠀⠀⠀⠀⠀⠀$\large{\bold I = \bold{ I_{1}} + \bold{ I_{2}} + \bold{ I_{3}}}$ ‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf{: \implies 10 + 5 + 4}$‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\sf\pink{ : \implies 19 A}$ ‎

Therefore the current flowing through each resistors is calculated to be:‎

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$\large\bold{: \implies I_{ 1 } = 10 A}$‎

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\bold{: \implies I_{ 2 } = 5 A}$‎

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\bold{ : \implies I_{ 3 } = 4 A}$‎

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The Total current is calculated to be,‎

$\sf\Large\red{ ⇝| I = 19 A |}$