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Up to now we have considered electric charges at rest. A flow of charge from one point to another is known as an electric current, and this normally takes place in a wire. If we measure the total charge flowing by any point in an interval of time, we have a measure of the current. We define current, i, to be:
i = Çq/Çt or current = charge/time The units for this new quantity are thus coulombs/second, or amperes (amps for short).
In many ways electric current is analogous to water flowing in pipes. It flows continuously without accumulating in any spot, and if its flow is reduced at any one point, the current in other parts of the circuits will be increased or decreased as needed to maintain continuity.
1. Current enters a junction as shown. What current leaves? (A is for ampere.)
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i = 2A + 7A i = 9A |
2. What are the unknown currents?
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13A = 4A + i1 i1 = 9A
i1 = 3A + i2 |
3. Find the missing currents.
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Just think about it:
i1 = 6A i2 = 3A i3 = 3A i4 = 8A |
In this text I have chosen to use the direction of electron flow as the direction of the current. I do this because in wires and resistors the moving charges are electrons. In ionized solutions and semiconductors this is not always the case, and the convention for electronics is that current is a positive charge flow (a decision made by Ben Franklin two centuries ago). The choice is really arbitrary for almost all circumstances, but I like to start out with the physical reality of fixed positive charge and moving electrons in a wire.A great deal of work has been done on the manipulation of current. By reducing its flow in one circuit element, current can be diverted to other elements and controlled to serve useful purposes. In the last century it was found that certain materials were more effective than others at reducing the flow of electricity that is, materials differ in their electrical resistivity.
Depending on its shape and the material or materials that compose it, every object has a measurable resistance to the flow of electric current. Look at some laboratory results using three different objects. We will hook them to a battery (a source of potential) and measure the resulting current that passes through each.
Here are the results, showing the current that results from different applied potentials. (We use E rather than V to represent potential. This is done in electronics, so we'll do it here. E should not be confused with E, the electric field.)
Note that in every case E «» i, so we may say that E = iR where R is a constant that depends on the particular object under study. We call R the resistance, and call the relationship Ohm's Law. The units of R are volts/amp or ohms, symbolized by the Greek letter capital omega: Ô.
4. A 12 V battery is connected to a 3 "omega" resistor. What current will flow through
the resistor?
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E = iR i = E/R i = (12 V)/(3Ô) i = 4.0 A |
5. A 3 A current flows through two resistors in series as shown. What is the potential
across the 6 "omega" resistor? What is it across the 8 Ô? What is it across the combined
resistance?
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E1 = iR E1 = (3 A)(6 Ô) E1 = 18 V
E2 = iR
E3 = E1 + E2 |
6. Find the current through each resistor:
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a) i = E/R i = (6 V)/(30 Ô) i = .20 A |
b) i = E/R i = (8 + 5)/500 i = .032 A |
c) i50 = E/R i50 = (6)/(50) i50 = .12 A
i100 = E/R |
7. The current through the circuit is 3 A. What is VAB, VBC,
VAC, VCD, and the potential of the battery?
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VAB = EAB VAB = iR VAB = (3)(7) VAB = 21 V
VBC = EBC
VAC = EAB + EBC
VCD = iR
E = VAC + VCD |
8. What potential does each battery have to produce the currents indicated?
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a) E = iR E = (.006)(3000) E = 18 V |
b) E = iR E = (2)(6) E = 12 V
E = iR note: the potential across both branches must be equal |
c) E40 = (.7)(40) E40 = 28 V
E60 = (.7)(60)
E = E40 + E60 |
9. Given EAB = -5 V and ECD = -3 V. Find EBE,
EEF, EAF, EBC, and EAC.
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Make a potential diagram, as show to help solve.
EBE = -7 V |
10. The current through the 5 Ô is 6 A.
a) What is the current through the 2 Ô?
b) What is the potential across the 8 Ô?
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a) i2 = i5 i2 = 6 A |
b) E8 = iR E8 = (6)(8) E8 = 48 V |
11. The potential across the 7 Ô is 21 V.
a) What is the current through the circuit?
b) What is the potential across the 4 Ô?
c) What is the potential of the battery?
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a) i = E/R i = (21)/(7) i = 3 A |
b) E4 = iR E4 = (3)(4) E4 = 12 V |
c) E = E4 + E7 E = 12 + 21 E = 33 V |
12. Find the current through the 16 Ô resistor.
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i = E/R i = (3)/(16) i = .188 A note: the 24 Ô resistor is irrelevant |
13. Find i1, i2, and i3.
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The full 20 V is applied across each resistor:
i4Ô = E/R
i8Ô = E/R
i5Ô = E/R
But:
i2 = i8Ô + i5Ô
i1 = i4Ô + i8Ô + i5Ô |
14. Find i1, i2, i3, and i4.
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i6Ô = (30)/(6)
i4Ô = (30)/(4)
i10Ô = (30)/(10)
i1 = i10Ô
i2 = i4Ô + i10Ô
i3 = i6Ô + i4Ô + i10Ô |
15. The current from the battery is 1.2 A. Find Eab, Ecd,
and Eae.
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The battery raises the potential to a high level. As the current flows through the
resistors, the potential drops in steps back to zero.
Eab = iR
Ebc = (1.2)(7)
Ecd = (1.2)(12)
Ede = (1.2)(4) Er = 31.2 V |
16. 6 A flows through the 12 ohm resistor. What current flows through the 7 ohm and
14 ohm resistors?
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The potential across the 7 Ô must equal the potential across the 14 Ô. Thus,
i1R1 = i2R2 or, in this case: (i7 Ô)(7) = (i14 Ô)(14) |
17. Find the unknown currents:
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i1 is 3 times i2 or 3/4 of the whole.
i1 = [(3)/(4)] * 10
i2 = [(1)/(4)] * 10 i3 is 9/(4 + 9) of the whole.
i3 = [(9)/(13)] * 6
i4 = [(4)/(13)] * 6 |
18. A current i enters two resistors as shown.
a) What is the potential across each resistor?
b) What is the total potential of the battery?
c) What is the effective resistance of R1 and R2 together?
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a) E1 = iR1 E2 = iR2 |
b) E = E1 + E2 E = iR1 + iR2 E = i(R1 + R2) |
c) R = E/i R = [i(R1 + R2)]/i R = R1 + R2 note: an unsurprising yet significant result |
19. Find the equivalent resistances:
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a) R = R1 + R2 R = 8 + 15 R = 23 Ô |
b) R = 200 + 50 + 120 R = 370 Ô |
c) R = 6 + 18 + 10 + 5 R = 39 Ô |
20. Find the current through the circuit:
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i = E/R i = (60)/(90 + 250 + 400) i = .0818 A |
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