Chapter 28: Q. 60 (page 793)

For the circuit shown in FIGURE $P 28.60$ , find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

Short Answer

Expert verified

Step by step solution

Given information

We have given that the circuit.

We need to find the current through and potential difference across each resistor.

Simplify

The given circuit could be simplified to be drawn as shown in the figure below. If we have two points and there resistors connected at both ends of the points, we called that the resistors are connected in parallel connection. Equation $(28.24)$ shows the equivalent resistance for the parallel connection in the form

$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + . . . + \frac{1}{R_{N}}$ $(1)$

The three resistors $4 Ω, 48 Ω$ and $16 Ω$ are in parallel, so we use equation $(1)$ to get their combination by

$\frac{1}{R_{1, e q}} = \frac{1}{4 Ω} + \frac{1}{16 Ω} + \frac{1}{48 Ω} R_{1, e q} = \frac{1}{4 Ω} + \frac{1}{16 Ω} + {\frac{1}{48 Ω}}^{- 1} R_{1, e q} = 3 Ω$

Simplify

fill

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For the circuit shown in FIGURE $P 28.60$ , find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

Short Answer

Step by step solution

Given information

Simplify

Simplify

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For the circuit shown in FIGURE P28.60, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.

Short Answer

Step by step solution

Given information

Simplify

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Famous Physicists

Energy Physics

Measurements

Fundamentals of Physics

Classical Mechanics

Turning Points in Physics

Study anywhere. Anytime. Across all devices.

For the circuit shown in FIGURE $P 28.60$ , find the current through and the potential difference across each resistor. Place your results in a table for ease of reading.