The circuit in Figure P 18.52 a consists of three resistors...

The circuit in Figure P 18.52 a consists of three resistors and one battery with no internal resistance. (a) Find the current in the $5.00-\Omega$ resistor. (b) Find the power delivered to the $5.00-\Omega$ resistor. (c) In each of the circuits in Figures $\mathrm{P} 18.52 \mathrm{b}$ $\mathrm{P} 18.52 \mathrm{c},$ and $\mathrm{P} 18.52 \mathrm{d},$ an additional $15.0-\mathrm{V}$ battery has been inserted into the circuit. Which diagram or diagrams represent a circuit that requires the use of Kirchhoff’s rules to find the currents? Explain why. (d) In which of these three new circuits is the smallest amount of power delivered to the $10.0-\Omega$ resistor? (You need not calculate the power in each circuit if you explain your answer.)

Key Concepts

Ohm's Law

Ohm's Law relates the voltage across a resistor to the current flowing through it and its resistance. This fundamental principle (V = IR) is essential for calculating currents and voltage drops in any circuit analysis involving resistors and power sources.

Kirchhoff's Circuit Laws

Kirchhoff's Circuit Laws, which include Kirchhoff's Voltage Law (the sum of voltage drops in a closed loop is zero) and Kirchhoff's Current Law (the total current entering a junction equals the total current leaving it), are crucial for analyzing complex circuits that cannot be simplified into single series or parallel combinations. These rules allow for establishing equations to solve for unknown currents and voltages in multi-loop circuits.

Series and Parallel Circuits

Understanding the configuration of resistors and sources, whether arranged in series or parallel, is key in circuit analysis. Series circuits sum their resistances, affecting the overall current, while parallel circuits share the voltage equally but have combined resistances calculated using reciprocal sums. This understanding helps in simplifying complex circuits and determining the behavior of the circuit elements.

Power Calculation in Electric Circuits

Power in electrical circuits is calculated using formulas such as P = IV, P = I^2R, or P = V^2/R. Calculating the power delivered or consumed by circuit elements is important for assessing circuit performance, energy distribution, and ensuring components operate within specified limits.

Transcript

00:01 In this particular problem, we need to find the current in the phi -oom register in the step a.

00:09 Now, we'll go step by step to start with step a.

00:18 What we need to find is current through phi -oom register.

00:21 So we'll give the notation as i -fi -oam that we need to find out.

00:28 Now, if you see figure a, now in this case, phi -oom and 8 -oom are in parallel.

00:39 So we'll write now in figure a, phi -oom and 8 -oom are in parallel combination.

01:03 Therefore, rp, that is parallel resistance, becomes 5 -oom and 8 -oom, are in parallel.

01:12 So we'll calculate by using formula as 5 -28 upon 5 plus 8, which is coming as 3 .076 -homes.

01:23 Which is the value of rp that is parallel combination of fiom and eight o now let's draw the diagram to understand this so the diagram will be this is the ten om which is there in figure a the parallel combination that we have got is of 3 .076 homes and that is connected with the battery which is of 15 volt so since 3 .0 076 oms and 10 om they are connected in series because the same current is flowing through them which is i so therefore using simple oms law i is equal to v upon r so v is of 15 volt and r will be a series combination of these two resistances 3 .076 plus ten oom so therefore the current which is flowing through this circuit is of 1 .147 amper let's mark this first.

02:38 Now, in the next page, what we are going to do is, so, now this is a continuation of step a only.

02:53 Now, so voltage drop, we'll write it down, voltage drop across 3 .076 home will be 3 .076 multiplied by 1 .147.

03:17 So resistance and the current which is flowing through it, which is going to be 3 .53 volt.

03:25 So 3 .53 volt is the voltage drop across 3 .076 oms.

03:33 Now since 3 .076 oms is in parallel combination of 5 oms and 8 oms, voltage across 3 .076 oms, voltage across 3 .3 .3 .5 oms, voltage.

04:02 Phi -oam is also 3 .53 volts.

04:09 This is what we need to understand since it was a parallel combination.

04:16 So ultimately the current in phi -oam which is equal to i -fi -oam therefore so current in phi -oom becomes 3 .53 volts divided by phi -oam so which is coming as 0 .706m so ultimately in order to write answer that is current through pho becomes 0 .706 amper.

04:50 This is what is the answer for step a.

04:55 This is the final answer...

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