Question

a 12V battery 6w electric bulb ammeter and plug key are connected in a circuit in series calculate the current flowing in the circuit and calculate the resistance of the bulb

Answer 1

P = VI
So 6= 12 × I
I = 0.5
Current is 0.5 A

V= IR
R = V/I
R= 12/0.5
R= 24 Ohm
I think it's correct now

Answer 2

Here is your answer...

$v = 12v \\ p = 6w \\ p = iv \\ i = \frac{p}{v} \\ i = \frac{6}{12} \\ i = \frac{1}{2} = 0.5a$

$v = ir \\ r = \frac{v}{i} \\ r = \frac{12}{ \frac{1}{2} } \\ r = 12 \times 2 \\ r = 24 \: ohms.$

Thanks..

hope it helps you