Q44P Question: Three identical light ... [FREE SOLUTION]

Chapter 18: Q44P (page 760)

Question: Three identical light bulbs are connected to two batteries as shown in Figure 18.106. (a) To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-trip path that begins and ends at the same location. Each segment of the path should go through a wire, a bulb, or a battery (not through the air). How many valid energy conservation (loop) equations is it possible to write for this circuit? (b) Which of the following equations are valid energy conservation (loop) equations for this circuit? $E_{1}$ refers to the electric field in bulb 1; $L$ refers to the length of a bulb filament. Assume that the electric field in the connecting wires is small enough to neglect.
(1) $+ E_{2} L - E_{3} L = 0$ , (2) $E_{1} L - E_{3} L = 0$ , (3) $+ 2 e m f - E_{2} L - E_{3} L = 0$ , (4) $E_{1} L - E_{2} L = 0$ , (5) $+ 2 e m f - E_{1} L - E_{2} L = 0$ , (6) $+ 2 e m f - E_{1} L - E_{3} L = 0$ , (7) $+ 2 e m f - E_{1} L - E_{2} L - E_{3} L = 0$ . (c) It is also necessary to write charge conservation equations (node) equations. Each such equation must relate electron current flowing into a node to electron current flowing out of a node. Which of the following are valid charge conservation equations for this circuit? (1) $i_{1} = i_{3}$ , (2) $i_{1} = i_{2}$ , (3) $i_{1} = i_{2} + i_{3}$ . Each battery has an emf of $1.5 V$ . The length of the tungsten filament in each bulb is $0.008 m$ . The radius of the filament is $5 脳 10^{- 6} m$ (it is very thin!). The electron mobility of tungsten is localid="1668588909714" $1.8 脳 10^{- 3} (m / s) / (V / m)$ . Tungsten has localid="1668588927161" $6 脳 10^{28}$ mobile electrons per cubic meter. Since there are three unknown quantities, we need three equations relating these quantities. Use any two valid energy conservation equations and one valid charge conservation equation to solve for localid="1668588943223" $E_{1}, E_{2}, i_{1}$ and localid="1668588965567" $i_{2}$ .

Short Answer

Expert verified

The energy conservation equations are $P_{0} - P_{1} - B_{1} - B_{2} - P_{2} - P_{4} - P_{0}$ , $P_{0} - P_{1} - B_{1} - B_{3} - P_{3} - P_{4} - P_{0}$ and $P_{2} - P_{3} - B_{3} - B_{2} - P_{0}$ .

Step by step solution

Write the given data from the question.

Emf of the battery, $V = 1.5 V$

The length of the tungsten filament, $L = 0.008 m$ .

Radius of filament, $r = 5 脳 10^{- 6} m$

Electron mobility of tungsten, $渭 = 1.8 脳 10^{- 3} (m / s) / (V / m)$

The number of mobile electrons, $N = 6 脳 10^{28} e / m^{3}$

Write the statements that helps to write the conservation equation.

The current is always flow from the higher potential to the lower potential and energy remains conserve.

Write the energy conservation equation that must involve a round-trip path that begins and ends at the same location.

Write the energy conservation equation.

The energy conservation equation for the loop that starts from $P_{0}$ and go through bulb $B_{1}$ and $B_{2}$ is given as follows.

$P_{0} - P_{1} - B_{1} - B_{2} - P_{2} - P_{4} - P_{0}$

The energy conservation equation for the loop that starts from $P_{0}$ and go through bulb $B_{1}$ and role="math" localid="1668588087308" $B_{3}$ is given as follows.

$P_{0} - P_{1} - B_{1} - B_{3} - P_{3} - P_{4} - P_{0}$

The energy conservation equation for the loop that starts from $P_{2}$ and go through bulb $B_{3}$ is given as follows.

$P_{2} - P_{3} - B_{3} - B_{2} = 0$

Hence the energy conservation equations are $P_{0} - P_{1} - B_{1} - B_{2} - P_{2} - P_{4} - P_{0}$ , $P_{0} - P_{1} - B_{1} - B_{3} - P_{3} - P_{4} - P_{0}$ and $P_{2} - P_{3} - B_{3} - B_{2} - P_{0}$ .