Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in parallel to the first two. The total power dissipated in this circuit is:

(A) 2P/3

(B) 3P

(C) 2P

(D) 3P/2

(E) P

The total power dissipated in this circuit is 3P/2

Explanation:

Given that the power of each bulb is P

Let R represents their resistance

From V = IR --- make I the subject of formula

I = V/R

and P = VI

where V = Voltage, I = Current, R = Resistance and P = Power

So, P = V²/R (1)

From the question, we understand that two bulbs are connected serially.

This means that they operate at the same voltage..

So, the total resistance for bulb 1 and 2 is R + R = 2R

Using I = V/R

For bulb 1 and 2

I = V/2R.

So, their power, P is calculated as

P1 = VI

P1 = V * V/2R

P1 = V²/2R

From (1) above, we noted that P = V²/R

So,

P1 = ½V²/R

P1 = ½P.

The power in bulb 1 and 2 is ½P.

The third build is connected in parallel to the first two bulbs.

So the total power is calculated as

½P + P

= P(½+1)

= P(3/2)

=3P/2

connect the batteries in series and hope

not too sure thinking all accelerations are pointed down