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Chapter 21: Q35PE (page 776)

Apply the junction rule at point a in Figure 21.52.

Short Answer

Expert verified

Using the junction rule on point 'a', we obtained:

I3=I2+I1

Step by step solution

01

Definition of Kirchhoff's law

Kirchhoff's first law defines currents at circuit junctions. It states that the sum of currents flowing into and out of a junction in an electrical circuit equals the sum of currents flowing out of the junction.

02

Explanation for the junction rule

The sum of currents entering the junction equals the sum of currents leaving the junction, according to the junction rule. An equation may be written for point an in the figure 21.55 using the following rule:

I3=I2+I1

Hence current at location 'a' can be calculated using Kirchhoff's Junction rule as follows:

I3=I2+I1

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Most popular questions from this chapter

(a) What is the terminal voltage of a large 1.54 v carbon-zinc dry cell used in a physics lab to supply 2.00 a to a circuit, if the cell's internal resistance is 0.100Ω ? (b) How much electrical power does the cell produce? (c) What power goes to its load?

Electric fish generate current with biological cells called electro-plaques, which are physiological emf devices. The electro-plaques in the South American eel are arranged in \(140\) rows, each row stretching horizontally along the body and each containing \(5000\) electro-plaques. Each electro-plaque has an emf of \(0.15{\rm{ }}V\) and internal resistance of \(0.25{\rm{ }}\Omega \). If the water surrounding the fish has resistance of \(800{\rm{ }}\Omega \), how much current can the eel produce in water from near its head to near its tail?

Suppose you measure the terminal voltage of a 3.200-Vlithium cell having an internal resistance of5.00Ωby placing a1.00-kΩvoltmeter across its terminals. (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.

Two different 12-Vautomobile batteries on a store shelf are rated at 600and 850“cold cranking amps.” Which has the smallest internal resistance?

Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements?
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