Question

I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!

Answer 1

Answer:

a)

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

• V = 12 V
• P = 24 W

$\implies \mathsf{24=\frac{12^2}{R} }$

$\implies \mathsf{24R=12^2 }$

$\implies \mathsf{24R=144 }$

=> R= 6 Ohms(Ω)

b)

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

• Power (P)= 100 W

these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,

• V = 240 V

$\implies \mathsf{100=\frac{240^2}{R} }$

R and 100 can interchange places

$\implies \mathsf{R=\frac{240^2}{100} }$

$\implies \mathsf{R=\frac{57600}{100} }$

=> R = 576 Ω

By Ohm's Law:

$\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}$

=> 240 = I × 576

=>

=> I = 0.417 A

c)

I don't know it's resistance,... so sorry

d)

The brightness of the bulb in series is less than when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series is more than the individual resistance.

And

Brightness, i. e., Power

$\boxed{\mathfrak{Power \propto \frac{1}{Resistance} }}$

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.