Question

how many bulbs each of resistance 32ohm should be joined in parallel to draw a current of 3A from a battery of 6volt​

Answer 1

Answer:

16 bulbs.

Explanation:

We have been given:

Potential difference = 6V

and Current = 3A

To find the resistance we need use Ohm's law,

V=IR

$\\\tt 6V = (3A)R \\\tt R = 2 \Omega$

So we must arrange 32 Ω-s in such a way that we get 2 Ω as net resistance.

Now using parallel net resistance relation,

$\\\tt \dfrac{1}{R_{net}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} . \ . \ . + \dfrac{1}{R_{n}}$

And if the resistances were same:

$\\\tt \dfrac{1}{R_{net}} = \dfrac{1}{R} + \dfrac{1}{R} . \ . \ . n \ resistances \\\\\tt \dfrac{1}{R_{net}} = n \times( \dfrac{1}{R} )\\\\\tt R_{net} = \dfrac{R}{n}$

Similarly,

$\\\tt \dfrac{1}{R_{net}} = \dfrac{1}{32} + \dfrac{1}{32} + . \ . \ . 16 \ similar \ resitances \\\\\tt \dfrac{1}{R_{net}} = 16 \times \dfrac{1}{32} = \dfrac{1}{2} \\\\\tt R_{net} = 2 \Omega$

Therefore 16 bulbs re needed.

Answer 2

Given :-

Resistance 32ohm should be joined in parallel to draw a current of 3A from a battery of 6volt​

To Find :-

Total Bulb

Solution :-

Let the number of bulb be x

So,

Resistance = 1/R1 ... 1/Rn

1/R = x/32

R = 32/x

Now

We know that

V = IR{Ohm's law}

6 = 3 × 32/x

6x = 3 × 32

6x = 96

x = 96/6

x = 16