91Ó°ÊÓ

In a series circuit, all components are connected end-to-end, forming a single path for current flow.

In a parallel circuit, all components are connected across each other, forming exactly two sets of electrically common points.

(a)

When the two bulbs are connected in series, the same current through each bulb. The equivalent resistor of the circuit; Since they are connected in series, the equivalent resistor is the sum of their resistances, therefore,

$$\begin{gathered} {\mathrm{R}}_{\mathrm{equivalent}}=400 \mathrm{Î\copyright }+800 \mathrm{Î\copyright } \\ =1200 \mathrm{Î\copyright } \end{gathered}$$

Form Ohm's law, the current flowing through each bulb is:

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{eq}}} \\ =\frac{120}{1200} \\ =0.100\mathrm{A} \\ {\mathrm{I}}_1={\mathrm{I}}_2=0.100\mathrm{A} \end{gathered}$$

Therefore the current through each bulb is 0.100 A

(b)

The power dissipated in a resistor R with current I and a potential difference across the resistor V is given by:

$$\begin{gathered} \mathrm{P}=\mathrm{VI} \\ \mathrm{P}={\mathrm{I}}^2\mathrm{R} \\ \mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}…………………….\left(1\right) \end{gathered}$$

Thus, the power dissipated in each bulb can be estimated as,

$$\begin{gathered} {\mathrm{P}}_1={\mathrm{I}}^2{\mathrm{R}}_1 \\ {\mathrm{P}}_1=(0.100\mathrm{A}{)}^2(400\mathrm{Î\copyright }) \\ {\mathrm{P}}_1=4.00\mathrm{W} \\ {\mathrm{P}}_2={{\mathrm{I}}_2}^2{\mathrm{R}}_2 \\ {\mathrm{P}}_2=(0.100\mathrm{A}{)}^2\left(800\mathrm{Î\copyright }\right) \\ {\mathrm{P}}_2=8.00\mathrm{W} \end{gathered}$$

Therefore the power dissipated in the bulbs is 4.00 W and 8.00 W

And therefore, the totalpowerdissipated in both bulbs is,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{total}}={\mathrm{P}}_1+{\mathrm{P}}_2 \\ {\mathrm{P}}_{\mathrm{total}}=4.00 \mathrm{W}+8.00 \mathrm{W} \\ {\mathrm{P}}_{\mathrm{total}}=12.0 \mathrm{W} \end{gathered}$$

Therefore the total power is 12.0 W

(d)

When the two bulbs are connected in parallel, the same voltage is across each bulb.

Therefore from Ohm's law, the current flowing through each bulb is,

$$\begin{gathered} {\mathrm{I}}_1=\frac{\mathrm{V}}{{\mathrm{R}}_1} \\ {\mathrm{I}}_1=\frac{120 \mathrm{V}}{400 \mathrm{Î\copyright }} \\ {\mathrm{I}}_1=0.300\mathrm{A} \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_2=\frac{\mathrm{V}}{{\mathrm{R}}_2} \\ {\mathrm{I}}_2=\frac{120 \mathrm{V}}{800 \mathrm{Î\copyright }} \\ {\mathrm{I}}_2=0.150 \mathrm{A} \end{gathered}$$

Therefore the current through each bulb when connected in parallel is 0.300 A and 0.150 A

(e)

Using equation (1), the power dissipated in each bulb is,

$$\begin{gathered} {\mathrm{P}}_1^{\text{'}}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_1} \\ {\mathrm{P}}_1^{\text{'}}=\frac{(120 \mathrm{V}{)}^2}{400 \mathrm{Î\copyright }} \\ {\mathrm{P}}_1^{\text{'}}=36 \mathrm{W} \\ {\mathrm{P}}_2^{\text{'}}=\frac{(120 \mathrm{V}{)}^2}{800 \mathrm{Î\copyright }} \\ {\mathrm{P}}_2^{\text{'}}=18\mathrm{W} \end{gathered}$$

Therefore the power in both the bulb connected in parallel is 36 W and 18 W

(f)

And so, the total power dissipated in both bulbs is,

$$\begin{gathered} {{\mathrm{P}}^{\text{'}}}_{\mathrm{total}}={\mathrm{P}}_1^{\text{'}}+{\mathrm{P}}_2^{\text{'}} \\ {{\mathrm{P}}^{\text{'}}}_{\mathrm{total}}=36 \mathrm{W}+18 \mathrm{W} \\ {{\mathrm{P}}^{\text{'}}}_{\mathrm{total}}=54 \mathrm{W} \end{gathered}$$

Therefore the total power is 54 W

(g)

In the case when the two bulbs are connected in series, the same current flows through each bulb, and according to equation (1),

$\mathrm{P}={\mathrm{I}}^2\mathrm{R}$

the power dissipated in the bulb with higher resistance is larger therefore the bulb with resistance $800\mathrm{Î\copyright }$glows brighter.

In case When the two bulbs are connected in parallel, the same voltage is across each bulb, and according to equation (1),

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

the power dissipated in the bulb with smaller resistance is larger; therefore the bulb with resistance$400\mathrm{Î\copyright }$ glows brighter.

(h)

From the results of (c) and (f), the situation in which the two bulbs are connected in parallel has the greater total light output from both bulbs combined-

This is expected because according to equation (1), the circuit with a higher current dissipates larger power.

And since the equivalent resistance in the parallel bulb situation is smaller than that in the series situation, the current is higher in the parallel situation, according to Ohm's law.

Therefore, the situation in which the bulbs are connected in parallel has a greater total light output from both bulbs