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WJSwanson
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This is the type of problem I've been told to expect for my circuits exam tomorrow, so if anyone can provide some guidance or criticism I would greatly appreciate it.
Given the circuit pictured below, find:
I. The size and direction of the current at C.
II. The equivalent resistance on loop ADBA.
III. The size and direction of the current across R3.
IV. The size and direction of the current across R3 if a new power source of E1 is added before point D.
Assume four significant figures.
Note: I'll use "E" for the emf, since I'm not sure how to enter a script "E" except for a lowercase epsilon, which would just be weird. <_<
The circuit is:
[itex]\Sigma i_{loop} = 0[/itex]
[itex]\Sigma\Delta V_{loop} = 0[/itex]
[itex]\Delta V_{R_{n}} = I_{n}/R_{n}[/itex]
[itex]I_{E} = E * R_{eq}[/itex]
I: We take the relevant loop to be ABCA. We traverse the loop counterclockwise beginning at emf 1. Because all elements on the loop are in series, we may take the current to be constant.
[itex]E_{1} - IR_{1} - E_{2} - IR_{1} = 0 \Rightarrow I = \frac{E_{1} - E_{2}}{2R_{1}} = \frac{5V}{2000\Omega} = 2.50 mA[/itex]
Because the returned value is positive, the current flows in the direction of our traversal (counterclockwise). This makes intuitive sense because it implies that the potential difference created by the 10V battery overpowers that of the 5V battery.
The current at point C is 2.50mA and its direction is counterclockwise.
II: The resistors on the loop can be broken into a pair of resistors in series and the parallel resistors R2 and R3 which can be taken as a single equivalent resistor whose resistance is given by a reciprocal summation:
[itex]R'_{eq} = 1 / \Sigma\frac{1}{R_{n}} = 1 / (\frac{1}{2000\Omega} + \frac{1}{1500\Omega}) = 857.1\Omega[/itex].
Because the parallel resistances have been resolved into a single resistance which is in series with the two other resistors of resistance R2, the total equivalent resistance is given by a direct summation:
[itex]R_{eq} = \Sigma R_{n} = 2R_{2} + R'_{eq} = 4000\Omega + 857.1\Omega = 4857\Omega[/itex].
III: The current through the entire equivalent resistance is given by
[itex]i = \frac{E_{2}}{R_{eq}} = \frac{5V}{4857\Omega} = 1.029 * 10^{-3}A[/itex]
and the potential at point D, the node which branches into parallel resistors of R3 = 1500Ω and R2 = 2000Ω, is given by
[itex]V_D = E_2 - 2iR_2 = 5 - 2 * 1.029 * 10^{-3}A * 2000\Omega = 0.8440V[/itex].
Since we know that the potential along the equipotential wire leading from the last junction back to E2 is zero, we can use the drop in potential over R3 to derive the current across it:
[itex]V = 0.8440 = \frac{i_3}{R_3} \Rightarrow i_3 = \frac{V}{R_3} = \frac{0.8440V}{1500\Omega} = 5.893 * 10^{-4}A = 0.5893mA[/itex]
and because the current through E2 has been established as being in the opposite direction of the battery's increasing potential (in part I), the current through loop ADBA flows clockwise. Thus, the current through R3 is 0.5893mA in the clockwise direction.
IV: When the additional voltage source is added, the Kirchhoff equation for the loop current (traversing it clockwise) becomes:
[itex]-E_2 + iR'_{eq} - E_3 + 2iR_2 = 0 \Rightarrow i = \frac{E_2 + E_3}{2R_2 + R'_{eq}} = \frac{15V}{4857\Omega} = 3.088mA[/itex]
if the new voltage source is aligned "cooperatively" with E2. If E3 is aligned opposite E2, the current is given by
[itex]i = \frac{E_2 - E_3}{2R_2 + R'_{eq}} = \frac{5V}{4857\Omega} = 0.5893mA[/itex].
In either case, the potential at D is given by
[itex]-5V + 2iR_2 \pm 10V + iR'_{eq} = 0 \Rightarrow iR'_{eq} = V_D = 5V \mp 10V - 2iR_2 = 5V \mp 10V - (3.088mA, 0.5893mA) * 4000\Omega = 5V \mp 10V - (12.35V, 2.357V)[/itex]
and as in step III the current over R3 is simply given by
[itex]i_3 = \frac{V_D}{R_3} = \frac{iR'_{eq}}{R_3}[/itex]
so long as we take point B and the wires connected to it to be our zero-potentials.
Homework Statement
Given the circuit pictured below, find:
I. The size and direction of the current at C.
II. The equivalent resistance on loop ADBA.
III. The size and direction of the current across R3.
IV. The size and direction of the current across R3 if a new power source of E1 is added before point D.
Assume four significant figures.
Note: I'll use "E" for the emf, since I'm not sure how to enter a script "E" except for a lowercase epsilon, which would just be weird. <_<
The circuit is:
Homework Equations
[itex]\Sigma i_{loop} = 0[/itex]
[itex]\Sigma\Delta V_{loop} = 0[/itex]
[itex]\Delta V_{R_{n}} = I_{n}/R_{n}[/itex]
[itex]I_{E} = E * R_{eq}[/itex]
The Attempt at a Solution
I: We take the relevant loop to be ABCA. We traverse the loop counterclockwise beginning at emf 1. Because all elements on the loop are in series, we may take the current to be constant.
[itex]E_{1} - IR_{1} - E_{2} - IR_{1} = 0 \Rightarrow I = \frac{E_{1} - E_{2}}{2R_{1}} = \frac{5V}{2000\Omega} = 2.50 mA[/itex]
Because the returned value is positive, the current flows in the direction of our traversal (counterclockwise). This makes intuitive sense because it implies that the potential difference created by the 10V battery overpowers that of the 5V battery.
The current at point C is 2.50mA and its direction is counterclockwise.
II: The resistors on the loop can be broken into a pair of resistors in series and the parallel resistors R2 and R3 which can be taken as a single equivalent resistor whose resistance is given by a reciprocal summation:
[itex]R'_{eq} = 1 / \Sigma\frac{1}{R_{n}} = 1 / (\frac{1}{2000\Omega} + \frac{1}{1500\Omega}) = 857.1\Omega[/itex].
Because the parallel resistances have been resolved into a single resistance which is in series with the two other resistors of resistance R2, the total equivalent resistance is given by a direct summation:
[itex]R_{eq} = \Sigma R_{n} = 2R_{2} + R'_{eq} = 4000\Omega + 857.1\Omega = 4857\Omega[/itex].
III: The current through the entire equivalent resistance is given by
[itex]i = \frac{E_{2}}{R_{eq}} = \frac{5V}{4857\Omega} = 1.029 * 10^{-3}A[/itex]
and the potential at point D, the node which branches into parallel resistors of R3 = 1500Ω and R2 = 2000Ω, is given by
[itex]V_D = E_2 - 2iR_2 = 5 - 2 * 1.029 * 10^{-3}A * 2000\Omega = 0.8440V[/itex].
Since we know that the potential along the equipotential wire leading from the last junction back to E2 is zero, we can use the drop in potential over R3 to derive the current across it:
[itex]V = 0.8440 = \frac{i_3}{R_3} \Rightarrow i_3 = \frac{V}{R_3} = \frac{0.8440V}{1500\Omega} = 5.893 * 10^{-4}A = 0.5893mA[/itex]
and because the current through E2 has been established as being in the opposite direction of the battery's increasing potential (in part I), the current through loop ADBA flows clockwise. Thus, the current through R3 is 0.5893mA in the clockwise direction.
IV: When the additional voltage source is added, the Kirchhoff equation for the loop current (traversing it clockwise) becomes:
[itex]-E_2 + iR'_{eq} - E_3 + 2iR_2 = 0 \Rightarrow i = \frac{E_2 + E_3}{2R_2 + R'_{eq}} = \frac{15V}{4857\Omega} = 3.088mA[/itex]
if the new voltage source is aligned "cooperatively" with E2. If E3 is aligned opposite E2, the current is given by
[itex]i = \frac{E_2 - E_3}{2R_2 + R'_{eq}} = \frac{5V}{4857\Omega} = 0.5893mA[/itex].
In either case, the potential at D is given by
[itex]-5V + 2iR_2 \pm 10V + iR'_{eq} = 0 \Rightarrow iR'_{eq} = V_D = 5V \mp 10V - 2iR_2 = 5V \mp 10V - (3.088mA, 0.5893mA) * 4000\Omega = 5V \mp 10V - (12.35V, 2.357V)[/itex]
and as in step III the current over R3 is simply given by
[itex]i_3 = \frac{V_D}{R_3} = \frac{iR'_{eq}}{R_3}[/itex]
so long as we take point B and the wires connected to it to be our zero-potentials.