91Ó°ÊÓ

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

(a)ItisclearthatcurrentthroughR1is-ThevoltageacrossR3,R4andR2isthesameanditisgivenby

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

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

Therefore the currenet in each bulb is 0.5A

(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

(c)WhenR4isremoved,theequivalentresistanceis

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

Therefore ethe resistance in each bulb is 6 ohms

Also, the current through R2 and R3 is

(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

(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 $R_1$