Question

In the circuit of Fig. E26.15, each resistor represents a light bulb. $\mathbf{R}\mathbf{1}\mathbf{}\mathbf{=}\mathbf{R}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{R}\mathbf{3}\mathbf{}\mathbf{=}\mathbf{R}\mathbf{4}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\Omega }$and E=9.00 Vm (a) Find the current in each bulb. (b) Find the power dissipated in each bulb. Which bulb or bulbs glow the brightest? (c) Bulb Ris now removed from the circuit, leaving a break in the wire at its R1, R2, R3position. Now what is the current in each of the remaining bulbs ? (d) With bulb R4 removed, what is the power dissipated in each of the remaining bulbs? (e) Which light bulb(s) glow brighter as a result of removing R4 ? Which bulb(s) glow less brightly?Discuss why there are different effects on different bulbs

Short answer

the currenet in each bulb is 0.5A

the power in each bulb is 10.2W and 1.12W

the resistance in each bulb is 6 ohms

the power in bulb is 8W and 2W

brighter is $R_2$and R3 and less bright is $R_1$

Step-by-step solution

Formula used

SinceR2,R3andR4areinparallel,sotheircombinationis
$R_{234}=\frac{1}{3}(4.5\Omega )=1.5\Omega$
Equivalent resistance of the circuit is
$R=R234+R1=1.5Q+4.5Q=6.0Q$

Step 2:Determine the current in each bulb

(a)ItisclearthatcurrentthroughR1is-ThevoltageacrossR3,R4andR2isthesameanditisgivenby

$$\begin{gathered} V_{ab},=ΗI{R}_1 \\ =9.0V—(1.5A)(4.5) \\ =2.25V \end{gathered}$$
Now let us use Ohm‘s law to get the current I2, I3 and I4 by

$\begin{array}{l}{l}_4,=l_2=l_3=\frac{v_{ab}}{4.5\Omega }\\ =0.5A\end{array}$

Therefore the currenet in each bulb is 0.5A

Step 3:Determine the power in each bulb 

(b) Since$$\begin{gathered} P=I^{2}R \\ and{R}_1=R_2=R_3=R_4 \end{gathered}$$

$$\begin{gathered} P_1=I_1^2{R}_1 \\ =\left(1.5\mathrm{A}{)}^2\right(4.5) \\ =10.12\mathrm{W} \\ {\mathrm{P}}_2={\mathrm{P}}_3={\mathrm{P}}_4=(0.5\mathrm{A}{)}^2\left(4.5\right) \\ =1.12\mathrm{W} \end{gathered}$$

therefore the power in each bulb is 10.2W and 1.12W

Determine the resistance in each bulb

(c)WhenR4isremoved,theequivalentresistanceis

$$\begin{gathered} {\mathrm{R}}_{\mathrm{ep}}={\mathrm{R}}_1+{\mathrm{R}}_{23} \\ =4.5\mathrm{\Omega }+\frac{1}{2}\left(4.5\mathrm{\Omega }\right) \\ =6.75\mathrm{\Omega } \\ {\mathrm{I}}_1=\frac{\mathrm{V}}{\mathrm{Req}} \\ =\frac{9}{6.75} \\ =1.33\mathrm{A} \\ {\mathrm{I}}_2={\mathrm{I}}_3=\frac{\widehat{\mathrm{I}}-{\mathrm{IR}}_1}{4.5} \\ =0.667\mathrm{A} \end{gathered}$$
$$\begin{gathered} R=R234+R1=1.5Q+4.5Q=6.0Q \\ R=R_{234}+R_1 \\ =1.5\Omega +4.5\Omega \\ =6\Omega \end{gathered}$$

Therefore ethe resistance in each bulb is 6 ohms

Also, the current through R2 and R3 is

Determine the ower of remaing bulb if 4th is reduced

(d) The power is directly proportional to the square current so$P=I^{2}R$ We can use the same steps in part (b) to get the

dissipated power for each bulb

$$\begin{gathered} P_1=I_1^2{R}_1 \\ =\left(1.33\mathrm{A}{)}^2\right(4.59) \\ =8\mathrm{W} \\ \mathrm{P}2=\mathrm{P}3= \\ =(0.667\mathrm{A}{)}^2\left(4.5\right) \\ =2\mathrm{W} \end{gathered}$$

Therefore the power in bulb is 8W and 2W

Step 6:Determine which light glow brighter 

(e) It is clear that $R_1$will be the brightest even after removing $R_4$- However, the brightness of $R_1$decreased, while that of $R_2$and $R_3$increased- This is because of removing $R_4$, increases the equivalent resistance of $R_2$and R3 from 1.5 Q to

2.25 9. Thus for a constant current, the potential difference across R2 and increases and while that of$R_1$ decreases

marginally.

So, brighter is $R_2$and R3 and less bright is $R_1$