Question

Light Bulbs in Series. A${\mathbf{P}}_{\mathbf{1}}\mathbf{=}\mathbf{60}\mathbf{W}\mathbf{}\mathbf{}\mathbf{120}\mathbf{V}$light bulb and a 200 W, 120Vlight bulb are connected in series across a line. Assume that the resistance of each bulb does not vary with current. (Note:This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a25W,120V light bulb dissipates when connected to a line.) (a) Find the current through the bulbs. (b) Find the power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why?

Short answer

the current through the bulb is 0.79A

the power dispatced is 142W and 42 W

the 60W bulb burn quickly

Step By Step Solution

Step-by-step solution

About the power dispatced in bulb

Power is the rate at which energy of any type is transferred; electric power is the rate at which electric energy is transferred in a circuit.In this section, we’ll learn not only what this means, but also what factors determine electric power

consider bulb

$$\begin{gathered} {\mathrm{V}}_1=120\mathrm{V} \\ {\mathrm{P}}_1=60\mathrm{W} \\ {\mathrm{V}}_2=120\mathrm{V} \\ {\mathrm{P}}_2=200\mathrm{W} \end{gathered}$$

Determine the Current through the bulbs

Solution
(a) have the voltage across each bulb, so we will the resistance of each bulb then use
Ohm‘s law to get the current ?ow$ in each bulb- The power of the bulb is related to its resistance by equation 25.18 in the
next form

$\mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}$

NowletuscalculateR1andR2
$$\begin{gathered} {\mathrm{R}}_1=\frac{{120}^2}{60} \\ =240\mathrm{\Omega } \\ {\mathrm{R}}_2=\frac{{120}^2}{200} \\ =72\mathrm{W} \end{gathered}$$

ThesecalculationsarebeforetheMobulbsareconnected-AfterconnectedinseriesacrossV=240Vthecurrentisth
same for both bulbs and equal to the current through their combination where the current could be calculated by
${\mathrm{l}}_1={\mathrm{l}}_2=\frac{\mathrm{V}}{{\mathrm{R}}_1+{\mathrm{R}}_2}$

ThesecalculationsarebeforetheMobulbsareconnected-AfterconnectedinseriesacrossV=240Vthecurrentisthe
same for both bulbs and equal to the current through their combination where the current could be calculated by

Now We can plug our values for V, R1 and R2 into equation (2) to get I1 and I2
$$\begin{gathered} {\mathrm{l}}_1={\mathrm{l}}_2=\frac{\mathrm{V}}{{\mathrm{R}}_1+{\mathrm{R}}_2} \\ =\frac{240}{240+72} \\ =0.79\mathrm{A} \end{gathered}$$

Therefore the current through the bulb is 0.79A

Determine the Power dispatced in each bulb 

(b)ThegivenpoWeristhepOWerwhenthebulbsoperatedbetWeen120V.30,togetthedissipatedpoWerbyeachbulb
when connected betWeen 240 V, We plug our values for V and R into equation (I) to get P for each bulb
$$\begin{gathered} {\mathrm{P}}_1={{\mathrm{I}}^2}_1{\mathrm{R}}_1=0.769{\mathrm{A}}^2\times 240 \\ =142\mathrm{W} \\ {\mathrm{P}}_2={{\mathrm{I}}_2}^2{\mathrm{R}}_2= \\ =0.769{\mathrm{A}}^2\times 72 \\ =42\mathrm{W} \end{gathered}$$

Therefore the power dispatced is 142W and 42 W

Step 3: Determine which bulb burns quickly

:c) The greatest power is dissipated by the 60 W bulb. So, it is heated very quickly and, therefore, burns quickly When :onnected to 240 V.

Therefore the 60W bulb burn quickly