Question

Question: Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of $\mathbf{400}\mathbf{\Omega }$and$\mathbf{800}\mathbf{\Omega }$. If the two light bulbs are connected in series across a$\mathbf{120}\mathbf{V}$line, find (a) the current through each bulb; (b) the power dissipated in each bulb; (c) the total power dissipated in both bulbs. The two light bulbs are now connected in parallel across the$\mathbf{120}\mathbf{V}$line. Find (d) the current through each bulb; (e) the power dissipated in each bulb; (f) the total power dissipated in both bulbs. (g) In each situation, which of the two bulbs glows the brightest? (h) In which situation is there a greater total light output from both bulbs combined?

Short answer

Answer

(a) The current through each bulb is 0.100A

(b) The power dissipated in each bulb is 4.00 W and 8.00 W, respectively.

(c) The total power dissipated through both bulbs is 12 W

(d)The current through each bulb when connected in parallel is 0.300A and 0.150A

(e) The power in both the bulb connected in parallel is 36 W and 18 W

(g) The total power is 54 W

(h) The situation in which the bulbs are connected in parallel has the greater total light output from both bulbs.

Step-by-step solution

 Step 1: About Series and Parallel connection

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.

Step 2:Deterimne the current through each bulb

(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,

$$\begin{gathered} R_{equivalent}=400\Omega +800\Omega \\ =1200\Omega \end{gathered}$$

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

$$\begin{gathered} I=\frac{V}{R_{eq}} \\ =\frac{120}{1200} \\ =0.100A \\ {I}_1=I_2=0.100A \end{gathered}$$

Therefore the current through each bulb is 0.100 A

Step 3:Determine the power dissipated in each bulb  

(b)

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

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

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

$$\begin{gathered} P_1=I^2{R}_1 \\ {P}_1={\left(0.100A\right)}^2\left(400\Omega \right) \\ {P}_1=4.00W \\ {P}_2={I_2}^2{R}_2 \\ {P}_2={\left(0.100A\right)}^2\left(800\Omega \right) \\ {P}_2=8.00W \end{gathered}$$

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

Determine the total power dissipated in both the bulb

(c)

And therefore, the totalpowerdissipated in both bulbs is,

$$\begin{gathered} P_{total}=P_1+P_2 \\ {P}_{total}=4.00W+8.00W \\ {P}_{total}=12.0W \end{gathered}$$

Therefore the total power is 12.0 W

 Step 5: Determine when two bulbs are connected in parallel

(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,

$$\begin{gathered} I_1=\frac{V}{R_1} \\ {I}_1=\frac{120V}{400\Omega } \\ {I}_1=0.300A \\ {I}_2=\frac{V}{R_2} \\ {I}_2=\frac{120V}{800\Omega } \\ {I}_2=0.150A \end{gathered}$$

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,

$$\begin{gathered} {P\text{'}}_1=\frac{V^2}{R_1} \\ {P\text{'}}_1=\frac{{\left(120V\right)}^2}{400\Omega } \\ {P\text{'}}_1=36W \\ {P\text{'}}_2=\frac{{\left(120V\right)}^2}{800\Omega } \\ {P\text{'}}_2=18 W \end{gathered}$$

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,

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

Therefore the total power is 54 W

Step 6:Determine the situation in which we get more power output  

(g)

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

$P=I^{2}R$

the power dissipated in the bulb with higher resistance is larger therefore the bulb with $800\Omega$resistance glows brighter.

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

$P=V^2/R$

the power dissipated in the bulb with smaller resistance is larger; therefore the bulb with $400\Omega$resistance 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