Question

How many bulbs of 8 ohm be joining in parallel to draw a current of 2A from a battery of 4V.

Answer 1

Here,

It is given that:

I = 2A and V = 4V 

Thn required resistance in the circuit,

R = >V/I

=> 4/2

=> 2 ohm.

Resistance of each resistor,

r = 8 ohm.

If n resistors, each of resistance r, are connected in parallel to get the required resistance R, then:

R = r/n

=> 2 = 8/n

=> n = 8/2

=> n = 4.

Therefore 4 resistors (bulb) should be joined in parallel to draw a current of 2A from a battery of 4V.

Answer 2

Heya!!

Here's your answer!!!

Current (i) = 2A
voltage (v) = 4v
residence (r) = ?

$r = \frac{v}{i} \\ \\ = r = \frac{4}{2}$

= r = 20 ohm

so the total resistance = 20ohm
assuming the no. of bulbs be = x

now,

$\frac{1}{re} = x \times \frac{1}{r1} \\ \\ = \frac{1}{2} \ = x \times \frac{1}{8}$

= x = 4

so it will required no. of bulbs = 4

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@vaibhavhoax
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