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General Physics II

Circuits

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Circuit Symbols

Circuit Diagram

• Workbook, Chapter 23: Question 2.

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Kirchhoff’s Laws• Junction law: Total current

entering a junction = total current leaving.

• Loop Law: Sum of potential differences around a closed loop = 0. Like gravitational potential energy, electric potential energy only depends on position, not path. Hence, if the starting and ending points are the same for a charge in a circuit, the overall change in potential energy is zero. Since ΔV = ΔU/q, the total change in potential is also zero.

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Applying Kirchhoff’s Laws• Choose a direction for the current if unknown. If the circuit is a

single loop, then the current is the same at all points.• Put plus and minus signs on the terminals of each emf source.• Apply the junction law if there are junctions in the circuit.• Choose a direction to move around a loop in order to apply the

loop law. • The change in potential across a resistor going in the direction of

the current is -IR. The change in potential across a resistor going in the opposite direction to the current is +IR.

• The change in potential in going across an emf source from negative terminal to positive terminal is + . The change in potential in going across an emf source from positive terminal to negative terminal is - .

• Apply the loop law. Solve for the unknown quantity (or quantities) using the equations obtained from the junction and loop laws.

E

E

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Applying Kirchhoff’s Laws

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Workbook, Chapter 23: Question 5.

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Series Circuits• In a series circuit, the circuit

elements are connected sequentially, i.e., each device is connected to its neighbor at only one end.

• Since the devices in series form a single unbroken path, the current is the same at all points.

• If the path is broken, e.g., by removing one bulb, the current ceases to flow.

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Resistors in Series• We wish to replace two resistors

connected in series by a single equivalent resistance.

• Using the loop law,1 2

1 2

0.( ).

IR IRI R R

− − == +EE

• But from the circuit with the equivalent resistor, the loop law yields

.eqIR=E

• It follows that

1 2.eqR R R= +

• For N resistors in series,

1 2 .... .eq NR R R R= + + +

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Workbook, Chapter 23: Question 8

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Parallel Circuits• In a parallel circuit, each

element is connected to the others at both ends.

• Each device has a separate pathto the battery. If one bulb is removed, the other one remains lit.

• For ideal (zero-resistance) wires, all points connected to the positive battery terminal are at the same potential (e.g., 1.5 V). Similarly, all points connected to the negative battery terminal are at the same potential (e.g., 0 V). Thus,

• all devices connected in parallel have the same potential difference across them.

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Resistors in Parallel• Using the Junction law,

1 2

1 2

1

2

2

1

1

12

1

. / ,

.

.

,

1 1 1 .1

batt

b

eq

att

batt eq

eq

eq

I I II V R

I R R

I

R R

R

R R R

R RR R

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

−

= +=Δ

= +

=

= +

= + ⇒ = +

E E

E

E E E

But, so

Now, from the equivalentcircuit, So

i.e.,

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Workbook: Chapter 23, Questions 7 and 9

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More Complex Circuits: Example 23.7

(1) The 600-Ω and 400-Ω resistors are in parallel:

• Break down circuit in steps, replacing combinations by their equivalent resistors

1

,11 1 240 .600 400 eqR

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

−= + = Ω

Ω Ω

(2) Req,1 and the 560-Ω are in series:

,2 ,1 240 800 .eq eqR R= + Ω= Ω

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Example 23.7(3) The equivalent resistor for the entire circuit replaces Req,2 and

the 800-Ω , which are in parallel:1

1 1 400 .800 800 eqR⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

−= + = Ω

Ω Ω

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Example 23.7

(1) The current leaving and entering the terminals of the battery is the current in the overall equivalent resistor. The potential difference across this resistor is the terminal voltage of the battery, which is equal to its emf.

• To find currents and potential differences, rebuild circuit byreversing the steps taken to find overall equivalent resistor.

12 V 0.030 A = 30 mA.400 batt

batt eq

VI R

Δ= = =

Ω

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Example 23.7

(2) Replace Req with the two parallel resistors it is equivalent to. Since the resistors are in parallel, the potential difference across each is equal to the potential difference across Req, which is 12 V. To find the currents, use ΔV = IR.

,2

800

12 V 0.015 A = 15 mA.800 12 V 0.015 A = 15 mA.800

eqRI

I Ω

= =Ω

= =Ω

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Example 23.7(3) Replace Req,2 with the series combination of Req,1 and the

560-Ω resistor. Since this is a series combination, the current in each resistor is the same as that in the equivalent resistor Req,2, which is 15 mA. Find the potential differences using ΔV = IR.

(4) Replace Req,1 with the parallel combination of the 600-Ωand 400-Ω resistors. Since the resistors are in parallel, the potential difference across each is equal to the potential difference across Req,1. To find the currents, use ΔV = IR.

• Workbook: Chapter 23, Question 13

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Chapter 23, Problem 59P23.59. Prepare: The figure shows how to simplify the circuit in Figure P23.59 using the laws of series and parallelresistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” thecircuit to find the current and potential difference of each resistor. We will assume that the battery and the connectingwires are ideal.

Solve: (a) From the last circuit in the figure and from Kirchhoff’s law, I = 100 V/10 Ω = 10 A. Thus, the current through the battery is 10 A. Now as we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference.

In Step 1 of the above diagram, we return the 10 Ω resistor to the 4.0 Ω, 4.0 Ω, and 2.0 Ω resistors in series. These resistors must have the same 10 A current as the 10 Ω resistance. That is, the current through the 2.0 Ω and the 4.0 Ω resistors is 10 A. The potential differences are

ΔV2 = (10 A)(2.0 Ω) = 20 V ΔV4 (left) = (10 A)(4.0 Ω) = 40 V ΔV4 (left) = (10 A)(4.0 Ω) = 40 V In Step 2, we return the left 4.0 Ω resistor to the 20 Ω and 5.0 Ω resistors in parallel. The two resistors must have the same potential difference ΔV = 40 V. From Ohm’s law,

I5 =40 V5.0 Ω

= 8.0 A I20 =40 V20 Ω

= 2.0 A

The currents through the various resistors are I2 = I4 = 10 A, I5 = 8.0 A, and I20 = 2.0 A. (b) The power dissipated by the 20 Ω resistor is I20

2 (20 Ω) = (2 A)2(20 Ω) = 80 W.