In the circuit of Fig. E26.15 each resistor represents a...

In the circuit of Fig. E26.15 each resistor represents a light bulb. Let $R_{1}=R_{2}=R_{3}=R_{4}=4.50 \Omega$ and $\mathcal{E}=$ 9.00 $\mathrm{V}$ . (a) Find the current in each bulb. (b) Find the power dissipated in each bulb.(b) Find the power dissipated in each bulb. Which bulb or bulbs glow the brightest? (c) Bulb $R_{4}$ is now removed from the circuit, leaving a break in the wire at its position. Now what is the current in each of the remaining bulbs $R_{1}, R_{2},$ and $R_{3} ?$ (d) With bulb $R_{4}$ removed, what is the power dissipated in each of the remaining bulbs? (e) Which light bulb(s) glow brighter as a result of removing $R_{4} ?$ Which bulb(s) glow less brightly? Discuss why there are different effects on different bulbs.

In the circuit of Fig. E26.15 each resistor represents a light bulb. Let $R_{1}=R_{2}=R_{3}=R_{4}=4.50 \Omega$ and $\mathcal{E}=$ 9.00 $\mathrm{V}$ . (a) Find the current in each bulb. (b) Find the power dissipated in each bulb.(b) Find the power dissipated in each bulb. Which bulb or bulbs glow the brightest? (c) Bulb $R_{4}$ is now removed from the circuit, leaving a break in the wire at its position. Now what is the current in each of the remaining bulbs $R_{1}, R_{2},$ and $R_{3} ?$ (d) With
bulb $R_{4}$ removed, what is the power dissipated in each of the remaining bulbs? (e) Which light bulb(s) glow brighter as a result of removing $R_{4} ?$ Which bulb(s) glow less brightly? Discuss why there are different effects on different bulbs.

Key Concepts

Ohm's Law

A fundamental principle in electrical circuits, Ohm's Law states that the voltage across a resistor is proportional to the current flowing through it, expressed as V = IR. This relationship is critical for calculating the current flows and voltage drops in various parts of a circuit, regardless of its complexity, and forms the basis for much of circuit analysis.

Series Circuits

In a series circuit, components are connected end-to-end so that there is only one path for current to flow. The total resistance in the circuit is the sum of the individual resistances, and the same current flows through each component. This concept helps in understanding how voltage divides among resistive elements and affects the brightness of bulbs modeled as resistors.

Parallel Circuits

A parallel circuit consists of branches where each component is connected directly to the voltage source, meaning all components experience the same voltage across them. However, the total current splits among the various branches according to their resistances. This concept is essential in analyzing circuits where multiple bulbs or resistors are connected across the same potential difference.

Current Division

Current division is a principle used to determine how the total current entering a parallel network divides among the individual branches. The current in each branch is inversely proportional to its resistance. This concept is particularly useful when analyzing circuits with multiple paths, such as when determining the brightness of each light bulb in the circuit based on the current it carries.

Power Dissipation in Resistive Circuits

Power dissipation in a resistor, which is often calculated using the formulas P = I²R or P = V²/R, represents the rate at which electrical energy is converted to heat or light. Understanding power dissipation is important for determining the relative brightness of light bulbs in a circuit, since the brightness correlates with the power consumed by the resistive elements.

Impact of Circuit Modifications (Open Circuit Effects)

Modifying a circuit by removing an element, such as creating an open circuit by removing one of the bulbs, changes the distribution of current and voltage among the remaining components. This concept involves analyzing how such changes affect both the overall circuit resistance and the performance of individual branches. Evaluating the effects of circuit modifications is key to predicting changes in brightness and power dissipation in the remaining bulbs.

Transcript

00:01 Hi friends as shown in the figure in the given circuit there are four electric length each off resistance r1 r2 r3 r4 4 .5 ome connected across the battery of emf 9 bolt in the first part we have to find the current through each valve we have to dip the power developed in each bulb see if r4 is removed then current through each bulb deep power in each bulb e part which bulb glow brighter let us see a first part in the first part the resistance r2 r3 and r4 are connected in parallel so equivalent resistance of 234 will be 4 .5 upon 3, that is 1 .5 o.

02:17 So above circuit can be reduced to this is r1 and this will be r234.

02:30 So current flowing through r1 and 234 will be e upon r equivalent that is 9 upon 1 .5 plus 4 .15 that is 1 .5 ampere.

02:50 Hence current through r1 is 1 .5 ampion and current through r2, r3 and r4 will be the same because they are connected in parallel and having same resistance.

03:07 So current can equally distributed between them.

03:10 So through each will be 0 .5 ampere.

03:14 Now we have to find the power to develop in the the second part, power is defined as ir1 square into r1 that is 1 .5 square into r1 into r1 that is 4 .5 so it is to be after calculating 8 power you will get 10 .1 power through r2, r3, r4 each having the same value 0 .5 square into 4 .5 square into 4 .5 that is 1 .1 2 watt.

04:14 Now see part.

04:15 If r4 is removed, then r2 and r3 are in parallel.

04:30 So it's an equivalent resistance you will get 4 .5 upon 2, that is 2 .25 o...

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