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Figure 1 2. (12 points, suggested time 25 minutes) The circuit shown in Figure ...

May 15, 2024

Solution

To compare the currents at points X, Y, and Z, we need to analyze the circuit. The current at point Z,

I_Z

, is the total current supplied by the battery. This current splits into two branches at point Z. The current through the top branch (with resistor

R_1

) is

I_{R1}

, and the current through the bottom branch (with resistors

R_3

and

R_4

in series) is

I_{R3+R4}

. Since the resistors in the bottom branch are in series, the current through

R_3

and

R_4

is the same,

I_{R3+R4}

. Therefore, the current at point Y,

I_Y

, is equal to

I_{R3+R4}

. The current at point X,

I_X

, is equal to

I_{R1}

. Using Kirchhoff's Current Law (KCL), we have:

I_Z = I_{R1} + I_{R3+R4}

Since

R_1 = 8R

R_3 = 3R

, and

R_4 = R

, the equivalent resistance of the bottom branch is:

R_{eq} = R_3 + R_4 = 3R + R = 4R

The current through each branch is inversely proportional to the resistance of the branch. Therefore, the current through the top branch is:

I_{R1} = \frac{I_Z \cdot R_{eq}}{R_1 + R_{eq}} = \frac{I_Z \cdot 4R}{8R + 4R} = \frac{I_Z \cdot 4R}{12R} = \frac{I_Z}{3}

The current through the bottom branch is:

I_{R3+R4} = \frac{I_Z \cdot R_1}{R_1 + R_{eq}} = \frac{I_Z \cdot 8R}{8R + 4R} = \frac{I_Z \cdot 8R}{12R} = \frac{2I_Z}{3}

Therefore, the currents at points X, Y, and Z are:

I_X = I_{R1} = \frac{I_Z}{3}

I_Y = I_{R3+R4} = \frac{2I_Z}{3}

I_Z = I_Z

Ranking the currents from greatest to least, we have: I_Z > I_Y > I_X So, the ranking is:

\mathrm{Z} \quad 1, \quad \mathrm{Y} \quad 2, \quad \mathrm{X} \quad 3

To determine the expressions for

I_X

and

I_Y

immediately after

t_X

and

t_Y

, we need to analyze the circuit in Figure 2. i. Immediately after

t_X

, the switch is closed to position X, and the capacitor starts charging. The current at position X,

I_X

, is given by:

I_X = \frac{\varepsilon}{R_1} = \frac{\varepsilon}{8R}

ii. Immediately after

t_Y

, the switch is closed to position Y, and the capacitor is fully charged. The current at position Y,

I_Y

, is given by:

I_Y = \frac{\varepsilon}{R_2 + R_3} = \frac{\varepsilon}{R + 3R} = \frac{\varepsilon}{4R}

To complete the bar chart for each resistor immediately after

t_X

and

t_Y

, we need to calculate the power dissipated by each resistor. Immediately after

t_X

: - Power dissipated by

R_1

P_1 = I_X^2 \cdot R_1 = \left(\frac{\varepsilon}{8R}\right)^2 \cdot 8R = \frac{\varepsilon^2}{64R} \cdot 8R = \frac{\varepsilon^2}{8R}

Immediately after

t_Y

: - Power dissipated by

R_2

P_2 = I_Y^2 \cdot R_2 = \left(\frac{\varepsilon}{4R}\right)^2 \cdot R = \frac{\varepsilon^2}{16R^2} \cdot R = \frac{\varepsilon^2}{16R}

- Power dissipated by

R_3

P_3 = I_Y^2 \cdot R_3 = \left(\frac{\varepsilon}{4R}\right)^2 \cdot 3R = \frac{\varepsilon^2}{16R^2} \cdot 3R = \frac{3\varepsilon^2}{16R}

The bar chart should reflect these power values relative to

P_1

d ⋮ To determine if the student's claim that

P_Y

is equal to

P_X

is correct, we need to compare the total instantaneous power dissipated by the circuit immediately after

t_X

and

t_Y

. Immediately after

t_X

P_X = P_1 = \frac{\varepsilon^2}{8R}

Immediately after

t_Y

P_Y = P_2 + P_3 = \frac{\varepsilon^2}{16R} + \frac{3\varepsilon^2}{16R} = \frac{4\varepsilon^2}{16R} = \frac{\varepsilon^2}{4R}

Comparing

P_X

and

P_Y

P_Y = \frac{\varepsilon^2}{4R} \neq P_X = \frac{\varepsilon^2}{8R}

Therefore, the student's claim is incorrect.

∻Answer∻ ⚹ The ranking of currents is: Z (1), Y (2), X (3). The expressions for $I_X$ and $I_Y$ are $\frac{\varepsilon}{8R}$ and $\frac{\varepsilon}{4R}$, respectively. The power dissipated by each resistor immediately after $t_X$ and $t_Y$ should be calculated and compared. The student's claim that $P_Y$ is equal to $P_X$ is incorrect. ⚹ ∻Key Concept∻ ⚹ Kirchhoff's Current Law (KCL) and Ohm's Law ⚹ ∻Explanation∻ ⚹ By analyzing the circuit using KCL and Ohm's Law, we can determine the current distribution and power dissipation in the circuit. ⚹