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How much current flows through the bottom wire in figureP28.66, and in which direction ?

Short Answer

Expert verified

The current flow through the bottom wire is0.42Ato the right direction.

Step by step solution

01

Given Information.

We need to find in which current flows in the bottom wire.

02

Explanation

We will apply the loop rule where the loop rule is a statement that the electrostatic force is conservative. Suppose we go around the loop, measuring potential differences across circuit elements as we go and the algebraic sum of these differences is zero when we return to the starting point.

V=0

Let us draw a loop clockwise and apply equation (28.2)for the left loop, where the voltage across the resistor 6Ωand 10Ωis negative because the traveling direction is in the direction of the current. The eemfε=9Vis positive because the direction of traveling is from negative to positive terminal in the battery.

V=09V-(6Ω)I6-(12Ω)I12=0I6=105V-2I12


Draw a loop clockwise and apply equation (28.2)for the middle loop, where the voltage across the resistor 12Ωis positive because the traveling direction is in the opposite direction of the current and 24Ωis negative because the traveling direction is in the direction of the current.

V=0-(24Ω)I24+(12Ω)I12=0I24=12I12/24I24=I122

03

Calculation

Draw a loop clockwise and apply equation (28.2)for the right loop, where the voltage across the resistor 10Ωand24Ωis negative because the traveling direction is in the direction of the current. The emfε=15Vis the positive because the direction of traveling is from negative to positive terminal in the battery.

V=015V-(10Ω)I10-(24Ω)I24=0I10=1.5V-2.4I24


From the junction rule, we get the current through the bottom wire Iwireat the left bottom junction by

Iwire=I6-I12=(1.5V-2I12)-I12=1.5V-3I12 (*)

So, for the junction rule at the right bottom junction, we get the current in the wire by

Iwire=I24-I10=I24-(105V-2.4I24)=-1.5V+3.4I24=1.5V+3.4(I122)=1.5V+1.7I12 (**)

Both previous equations have the same left side, so we can get I12value by,

-1.5V+1.7I12=105V-3I12I12=0.63A

Now, we plug the value for I12into equation (*) to get Iwireby,

Iwire=1.5V-3I12=1.5V-3(0.64A)=-0.42A

The negative sign indicates that the current travel to the right.

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Most popular questions from this chapter

An oscillator circuit is important to many applications. A simple oscillator circuit can be built by adding a neon gas tube to an RC circuit, as shown in figureCP28.83. Gas is normally a good insulator, and the resistance of the gas tube is essentially infinite when the light is off. This allows the capacitor to charge. When the capacitor voltage reaches a value Von, the electric field inside the tube becomes strong enough to ionize the neon gas. Visually, the tube lights with an orange glow. Electrically, the ionization of the gas provides a very-low-resistance path through the tube. The capacitor very rapidly (we can think of it as instantaneously) discharges through the tube and the capacitor voltage drops. When the capacitor voltage has dropped to a value Voff, the electric field inside the tube becomes too weak to sustain the ionization and the neon light turns off. The capacitor then starts to charge again. The capacitor voltage oscillates between Voff, when it starts charging, and Von, when the light comes on to discharge it.

a. Show that the oscillation period is

T=RCinε-Voffε-Von

b. A neon gas tube has Von=80VandVoff=20V. What resistor value should you choose to go with a 10μfcapacitor and a 90Vbattery to make a 10Hzoscillator?

You’ve made the finals of the Science Olympics! As one of your tasks, you’re given 1.0 g of aluminum and asked to make a wire, using all the aluminum, that will dissipate 7.5 W when connected to a 1.5 V battery. What length and diameter will you choose for your wire?

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.

What value resistor will discharge a 1.0μFcapacitor to 10% of its initial charge in 2.0ms?

Compared to an ideal battery, by what percentage does the battery’s internal resistance reduce the potential difference across the 20Ωresistor in FIGURE EX28.20?

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