Question

how many 12ohms resistors(in parallel) are required to carry 2A on 4V battery

Answer 1

When "n" resistors are connected in parallel ,

$\bigstar\ \; \sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_2}+...+\dfrac{1}{R_n}$

When "n" resistors are equal ,

⇒ R₁ = R₂ = R₃ = ... = Rn = R

$\bigstar\ \; \sf \dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+...+\dfrac{1}{R}\\\\\bigstar\ \; \sf{\dfrac{1}{R_{eq}}=\dfrac{n}{R}}\\\\\bigstar\ \; \sf \pink{R_{eq}=\dfrac{R}{n}}$

Given ,

Resistance , R = 12 Ω

Current , I = 2 A

P.d , V = 4 V

Apply Ohms Law ,

$\sf \to V=IR_{eq}\\\\\to \sf V=I \left( \dfrac{R}{n}\right)\\\\\to \sf 4=2\left( \dfrac{12}{n}\right)\\\\\to \sf 2=\dfrac{12}{n}\\\\\to \sf \orange{n=6\ }\ \; \bigstar$

Answer 2

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SOLUTION:-

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• Suppose you join 3 resistor in parallel thn

• 1/R=1/12+1/12+1/12
• 1/R-3/12-1/4

• R-4

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Similarly:-

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• If you add x resistor in parallel then

• 1/R-1/12+1/12+1/12..x times

• So you see above 3 resister join then

• 1/R-3/12
• Similarly for x resistors(x denote no. of
• resistor join in parallel)

• 1/R=x/12

• R=12/x

• You know current,voltage and resistance

• Then ohm's law

• V=IR

• V-4

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• R-12/x

• Put it 4-2*12/x =6

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