Question

Two light bulbs have constant resistances of 400 $\mathrm{\Omega }$ and 800 $\mathrm{\Omega }$. If the two light bulbs are connected in series across a 120-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 120-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

(a) 0.1 A; (b) 4 W and 8 W; (c) 12 W; (d) 0.3 A and 0.15 A; (e) 36 W and 18 W; (f) 54 W; (g) 800 Ω in series, 400 Ω in parallel; (h) Parallel.

Step-by-step solution

Identify Total Resistance in Series

When bulbs are connected in series, the total resistance, $R_{total}$, is the sum of their resistances. $$R_{total}=R_1+R_2=400\mathrm{\Omega }+800\mathrm{\Omega }=1200\mathrm{\Omega }$$

Calculate Current in Series

Using Ohm's Law, $V=IR$, calculate the current, $I_{series}$, through each bulb when connected in series. Here, $V=120\text{ V}$.$$I_{series}=\frac{V}{R_{total}}=\frac{120}{1200}=0.1\text{ A}$$

Calculate Power Dissipated in Each Bulb in Series

The power dissipated in any bulb is $P=I^{2}R$. Calculate for each bulb:$$P_1=(0.1{)}^2\times 400=4\text{ W}$$$$P_2=(0.1{)}^2\times 800=8\text{ W}$$

Calculate Total Power Dissipated in Series

The total power dissipated in bulbs connected in series is the sum of powers of each bulb:$$P_{total,series}=4+8=12\text{ W}$$

Identify Individual Resistance in Parallel

For parallel circuits, each bulb maintains its resistance: $R_1=400\mathrm{\Omega }$, $R_2=800\mathrm{\Omega }$.

Calculate Current Through Each Bulb in Parallel

Using Ohm's Law for parallel circuits, where the voltage across each is 120 V, compute the current. $$I_1=\frac{V}{R_1}=\frac{120}{400}=0.3\text{ A}$$$$I_2=\frac{V}{R_2}=\frac{120}{800}=0.15\text{ A}$$

Calculate Power Dissipated in Each Bulb in Parallel

Calculate power dissipated using $P=IV$ for each bulb: $$P_1=120\times 0.3=36\text{ W}$$$$P_2=120\times 0.15=18\text{ W}$$

Calculate Total Power Dissipated in Parallel

The total power dissipated in the parallel arrangement is the sum:$$P_{total,parallel}=36+18=54\text{ W}$$

Determine Which Bulb Glows Brightest in Each Sequence

Brightness is proportional to power dissipation. In series, the 800 Ω bulb (8 W) glows brighter than the 400 Ω bulb (4 W). In parallel, the 400 Ω bulb (36 W) glows brighter than the 800 Ω bulb (18 W).

Compare Total Light Output in Each Situation

Total light output is higher in parallel (54 W) compared to series (12 W), so the parallel configuration has greater light output.

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in the field of electrical engineering and physics. It is represented by the formula $V=IR$, which describes the relationship between voltage ($V$), current ($I$), and resistance ($R$). This law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them.

Understanding Ohm's Law is crucial for analyzing electrical circuits. It helps determine how much current will flow through a circuit for a given voltage and resistance. In the given problem, Ohm's Law is used to calculate the current flowing through two light bulbs when they are arranged in series and parallel configurations.

Remember that if you change one component in a circuit, such as increasing the resistance, it will affect the current flowing through the circuit. By using Ohm's Law, you can predict these changes and make informed decisions about modifying your circuit.

Series Circuits

A series circuit is a type of electrical circuit in which components are connected end-to-end in a single path. In this configuration, the current flowing through each component is the same, but the voltage across each component can differ.

To find the total resistance in a series circuit, simply sum up the resistances of all components:

• Total Resistance: $R_{total}=R_1+R_2+...+R_n$

In the provided exercise, two light bulbs with resistances of 400 Ω and 800 Ω are connected in series. The total resistance is 1200 Ω.

Using Ohm's Law, the current through each bulb is calculated by dividing the total voltage by the total resistance:

• Current through each bulb: 0.1 A

Since the current is the same for all components in a series circuit, both bulbs will have the same current flowing through them.

Parallel Circuits

Parallel circuits feature components connected with their ends joined across the same two points, creating multiple paths for current flow. The voltage across each component in a parallel circuit is the same.

In parallel circuits, the total resistance can be determined using:

• 1/Total Resistance: $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

For the exercise case, each bulb retains its individual resistance, but the voltage across each is 120 V.

Applying Ohm's Law to each bulb allows calculation of the current through each:

• Current through 400 Ω bulb: 0.3 A
• Current through 800 Ω bulb: 0.15 A

Parallel circuits allow different amounts of current to flow through different components, which can affect their power consumption and brightness.

Electrical Power

Electrical power is a measure of the rate at which electrical energy is consumed by a device. It is mathematically expressed as $P=IV$, where $P$ is power in watts, $I$ is current in amperes, and $V$ is voltage in volts.

Power dissipation in electric circuits can also be calculated using $P=I^{2}R$ or $P=\frac{V^2}{R}$. In the exercise, the power dissipated in each bulb is calculated when they are in series and parallel configurations:

For series:

• 400 Ω bulb dissipates 4 W
• 800 Ω bulb dissipates 8 W

Total power in series is 12 W, showing that series circuits divide voltage across components, impacting the total power output.

For parallel:

• 400 Ω bulb dissipates 36 W
• 800 Ω bulb dissipates 18 W

Total power in parallel is 54 W, demonstrating that parallel circuits provide higher combined power, resulting in greater brightness from the bulbs.