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Chapter 25: Problem 67

Two resistors with resistances \(200 . \Omega\) and \(400 . \Omega\) are connected (a) in series and (b) in parallel with an ideal 9.00-V battery. Compare the power delivered to the \(200 .-\Omega\) resistor.

Short Answer

Expert verified
Answer: The power delivered to the \(200\Omega\) resistor when connected in series is \(0.045W\), and the power delivered when connected in parallel is \(0.405W\).

Step by step solution

01

Find the Total Resistance in Series Connection

For resistors in series, the total resistance is the sum of the individual resistances. So, \(R_{total} = R_1 + R_2 = 200\Omega + 400\Omega = 600\Omega\).
02

Find the Current in Series Connection

Use Ohm's law, \(V = IR\), to find the current flowing through the circuit. Rearranging the equation and solving for the current gives \(I = \frac{V}{R_{total}} = \frac{9V}{600\Omega} = 0.015A\).
03

Calculate the Power Delivered to the \(200\Omega\) Resistor in Series Connection

We can now calculate the power delivered to the \(200\Omega\) resistor using the formula \(P = I^2R\). For the series connection, the current through both resistors is the same. Thus, \(P_{series} = (0.015A)^2(200\Omega) = 0.045W\).
04

Find the Equivalent Resistance in Parallel Connection

For resistors in parallel, the equivalent resistance is given by the formula \(\frac{1}{R_{equiv}} = \frac{1}{R_1} + \frac{1}{R_2}\). Plugging in the values, we get \(\frac{1}{R_{equiv}} = \frac{1}{200\Omega} + \frac{1}{400\Omega}\), which gives us \(R_{equiv} = 133.33\Omega\).
05

Find the Total Current in Parallel Connection

Again, use Ohm's law to find the total current flowing through the circuit. In this case, \(I_{total} = \frac{V}{R_{equiv}} = \frac{9V}{133.33\Omega} = 0.0675A\).
06

Find the Current through the \(200\Omega\) Resistor in Parallel Connection

We can find the current through the \(200\Omega\) resistor in a parallel connection using the current divider rule, which is as follows: \(I_{200} = I_{total}\frac{R_2}{R_1 + R_2} = 0.0675A \frac{400\Omega}{200\Omega +400\Omega} = 0.045A\).
07

Calculate the Power Delivered to the \(200\Omega\) Resistor in Parallel Connection

Now, we can calculate the power delivered to the \(200\Omega\) resistor using the formula \(P = I^2R\). For the parallel connection, \(P_{parallel} = (0.045A)^2(200\Omega) = 0.405W\). So, the power delivered to the \(200\Omega\) resistor when connected in series with the \(400\Omega\) resistor is \(0.045W\), and the power delivered when connected in parallel is \(0.405W\).

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