Question

Two 40 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb

Answer 1

Answer:

The power dissipated by each bulb is  $P = 10.0 \ W$

Explanation:

From the question we are told that

    The  power rating of both bulbs is  $P = 40 \ W$

     The voltage  rating of both bulb is  $V = 120 \ V$

     The  both bulbs are connected a voltage of  $V_C = 120 V$

The amount of power rating of each  bulb is mathematically represented as

       $P = \frac{V^2}{R }$

=>    $R = \frac{V^2}{P}$

substituting values

       $R = \frac{ (120)^2}{40}$

      $R = 360 \Omega$

Now given that the bulbs are connected is  series, the equivalent resistance is  evaluated as

          $R_{eq } = R +R$

substituting values

          $R_{eq } = 360 + 360$

         $R_{eq } =720 \ \Omega$

The  current flowing through the bulbs is mathematically evaluated as

         $I =\frac{V_C}{R_{eq}}$

substituting values

      $I =\frac{120}{720}$

      $I = 0.1667 \ A$

Now  the power dissipated by both bulbs is mathematically represented as

           $P = I ^2 * R$

substituting values      

         $P = 0.1668^2 * 360$

         $P = 10.0 \ W$

Answer 2

The power that should be dissipated by each bulb is P = 10.0 W.

Calculation of the power:

Since

The power rating of both bulbs is P = 40 W.

The voltage rating of both bulbs is V = 120 V.

And, both bulks that should be connected a voltage of Vc = 120V

Now the amount of power that should be rated of each bulb should be

P = V^2/R

So, R = V^2/P

= 120^2/40

= 360Ω

The equivalent resistance should be

I = Vc/Req

= 120/720

= 0.1667 A

Now the power is = 0.1668^2 * 360

= 10.0 W

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