e2

3 PAIR BARS What are the charges on plates 3 and 6?

+Q and −Q

milli micro nano pica

-3 -6 -9 -12

At $0.095/kWh, what does it cost to leave a 28 W porch light on day and night for a year?

.028 x .095 x 365 x 24

Suppose the electric company charges 10 cents per kWh. How much does it cost to use a 125 watt lamp 4 hours a day for 30 days?

.125 x 4 x 24 then times 10cents

The voltage through a circuit is v = 0.8 sin 314t. What is the amplitude (peak) of the voltage?

.8 V

3 PAIR BARS If the voltage across the first capacitor (the one with capacitance C) is ΔV1, then what are the voltages across the second and third capacitors?

1/2ΔV1 and 1/3ΔV1

A 3.00-μF and a 4.00-μF capacitor are connected in series, and this combination is connected in parallel with a 2.00-μF capacitor (Figure 1). What is the net capacitance?

1/3 + 1/4 7/12 12/7+2=3.7 uF

Two resistors having resistances of 5.0 Ω and 9.0 Ω are connected in parallel. A 4.0-Ω resistor is then connected in series with the parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination. What is the current through the 9.0-Ω resistor?

1/Rp flip then 3.2 + 4=7.2 6V/7.2 is I=.83A then half since parallel

Eight identical bulbs are connected in series across a 120-V line. What is the voltage across each bulb?

120/8=15V

A 600 −Ω and a 2800 −Ω resistor are connected in series with a 12-V battery. What is the voltage across the 2800 −Ω resistor?

12V x 2800/3400= 9.9V

Eight bulbs are connected in parallel to a 220 V source by two long leads of total resistance 1.9 Ω If 270 mA flows through each bulb, what is the resistance of each?

220/ (8x .27) then minus leads so 1.9

Resistances of 2, 4, and 6 Ώ and a 24-V emf device are all in parallel. The current in the 2Ώ resistor is:

24/2=12 A

PARALLEL 6 bars If the charge of the first capacitor (the one with capacitance C) is Q, then what are the charges of the second and third capacitors?

2Q and 3Q

1. In the diagram, the current in the 3Ω resistor is 4A. The potential difference between points 1 and 2 is:

4A (3+2)=20V

A 3.0-Ω resistor is connected in parallel with a 6.0-Ω resistor. This combination is then connected in series with a 4.0-Ω resistor. The resistors are connected across an ideal 12-volt battery. How much power is dissipated in the 3.0-Ω resistor?

5.3 W

Using the value of Qtot, find the equivalent capacitance Ceq for this combination of capacitors.

6C

Using the value of Q just calculated, find the equivalent capacitance Ceq for this combination of capacitors in series.

6C/11

Find the voltage ΔV1 across the first capacitor.

6ΔV/11

If the current is 0.52 A , what is the resistance of each bulb?

8R=V/I => R=(120/0.52)x(1/8) => R=28.8 Ohm

Rank below the three identical resistors (A, B, and C) in (Figure 1) on the basis of the current that flows through them. A B C

A alone is strong and B and C are even!

Which of the following will increase the resistance of a wire?

Decreasing the cross-sectional area of the wire will increase the resistance of the wire. Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire. Increasing the length of the wire will increase the resistance of the wire.

A 25 −W , 110-V bulb is connected in parallel with a 60 −W , 110-V bulb.

P = IV = V^2/R R=V^2/P 110^2 / 58= 142 omega

If the current is 0.52 A , what is the power dissipated in each bulb?

P=I2xR=0.52x28.8 => I=6.875 W

120V, 600W

P=IV 600/120

What is the rate at which thermal energy appears in R

P=I^2/R

If 270 mA flows through each bulb, what fraction of the total power is wasted in the leads?

Power in resistance of leads = (I^2)*Rleads 1.9/ last answer!!

A 100 W radiant heater is constructed to operate at 115 V. How much thermal energy is produced in 4 h?

Pt 100w x 4x 60x60

Find the charge Q on the first capacitor.

Q = CΔV1

PARALLEL 6 bars Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.

Qtot = 6CV

parallel

R1 and R3

Determine the magnitude of the current through R1 in the figure.(Figure 1)

R1 is around so add V (9+6) = 15/22=.68A V/R LEFT

What is the resistance of a circular rod, 1cm in diameter and 45m long, if the resistivity is 1.4x10-8 Ωm.

R= p (L/A) A= pir^2 r in m!!

Properties of resistors in series vs resistors in parallel.

Resistors in series have the same current and different voltages Resistors in parallel have the same voltage and different currents Capacitors in series have the same charge and different voltages Capacitors in parallel have the same voltage and different charges

The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a steady state.

charge

Rank below the three identical resistors (A, B, and C) in (Figure 1) on the basis of the current that flows through them.

completely alone is stronger compare to along then parallel!

When a thin copper wire that is 178 m long is connected between a 1.2-V potential difference, a current of 2.0 amps flows through the wire. What is the diameter of this wire? The resistivity of copper is 1.72 × 10-8 Ω • m.

d= sqrt(4pt/piR)

What is the value of the electric current?

e2-e1-ir-ir-ir2=0 i=e2-e1/r1+r2+r v/omega

Ammeter measures

electric current when is in series in a circuit.

A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what is the value of the current?

i= Q/t 1200/ (4x60)

At the same temperature, two wires made of pure copper have different resistances. The same voltage is applied at the ends of each wire. The wires may differ in

length. cross-sectional area. amount of electric current passing through them.

series

not a even cut!!

What is the minimum resistance you can obtain by combining these?

parallel 1/ (1/570)+(1/950)+(1/1200)

voltage unit

potential difference

Voltmeter measures

potential difference (voltage) when is connected in parallel (across) the resistor, battery or the 2points of interest from the circuit.

Calculate the equivalent resistance. All resistors are identical, R=3Ω:

remember flip!! 1/R

Ohmmeter measures

resistance when in parallel with the resistor

Calculate the equivalent capacitance

right angle 10 +10 1/27 + 1/27 + 1/20 20 x 27 /67

When different resistors are connected in parallel

the potential difference across each one is the same

In the circuit shown in the figure, R1 = 10 Ω, R2 = 12 Ω, R3 = 20 Ω, V1 = 1.0 V, V2 = 7.0 V, and the batteries are both ideal. What is the current through R1?

v2-v1/r1 6/10=.6V v2/r2+r3 7/32 = .2 .8A

Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.

Σ(ΔV)=0 = I3R3−I2R2

Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.

Σ(ΔV)=0 = Vb−I1R1−I3R3

Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2)

ΣI=0 = I2+I3−I1

Explain why an ideal ammeter would have zero resistance and an ideal voltmeter infinite resistance.

An ammeter measures the current it passes through it, therefore, it must offer zero resistance to be able to measure it, otherwise, it will generate a voltage drop when the current goes through it and it would need to be taken into account in the Kirchoff Law. An ammeter must always be connected in series, and if there is a voltage drop across the ammeter, the current would be modified due to the voltage drip across it. The resistance must be zero cross a ammeter in order for it to be able to correctly measure the value of the current without modifying it. For a voltmeter, the reason is similar. A voltmeter must always be connected in parallel and if there is not infinite resistance across the ideal voltmeter, current would flow through it.That is, the current that initially reached the resistance R in the diagram, would be decreased, thus modifying the voltage drop across it. The resistance across the voltmeter must be infinite, so no current flows through it and that way it can correctly measure the voltage difference between two points of the circuit!

Kirchhoff's junction rule:

The algebraic sum of the currents into (or out of) any junction in the circuit is zero.

A constant voltage is applied across a circuit. If the resistance in the circuit is doubled, what is the effect on the power dissipated by the circuit?

The power dissipated is reduced by a factor of 2.

Kirchhoff's loop rule:

The sum of the voltage changes across the circuit elements forming any closed loop is zero

Three 1.50-kΩ resistors can be connected together in four different ways, making combinations of series and/or parallel circuits What are these four ways?

Two resistors in series can be placed in parallel with the third. Two resistors in parallel can be placed in series with the third. The resistors can all be connected in parallel. The resistors can all be connected in series.

PARALLEL 6 bars If the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negative terminal of the battery is at zero potential.

V and 0

In the circuit shown in the figure (Figure 1), the 33−Ω resistor dissipates 0.60 W . What is the battery voltage?

V=IR W = I^2*R, so I = √[W/R] =sqrt .6/33 R is 33!! 33 + (68*75)/(68+75) =*just battery*

Select the expressions that will be equal to the voltage of the battery in the circuit, where VA, for example, is the potential drop across resistor A

VA+VB VA+VC VD

Suppose that you have a 570 −Ω , a 950 −Ω , and a 1.2 −kΩ resistor. What is the maximum resistance you can obtain by combining these?

add all **units!!!! change: 570+950+1200 =2700 omega

Current is a measure of:

amount of charge that moves past a point per unit time