week 7chapter 32

Fundamentals of Circuits

Resistors: devices that are pieces of conductorsmade to have specific resistance

Which of these diagrams represent the same circuit?

1. a and b

2. a and c

3. b and c

4. a, b, and c

5. a, b, and d

Which of these diagrams represent the same circuit?

1. a and b

2. a and c

3. b and c

4. a, b, and c

5. a, b, and d

What is ∆V across the unspecified circuit element? Does the potential increase or decrease when traveling through this element in the direction assigned to I?

1. ∆V increases by 2V in the direction of I.

2. ∆V decreases by 2V in the direction of I.

3. ∆V increases by 10V in the direction of I.

4. ∆V decreases by 10V in the direction of I.

5. ∆V increases by 26V in the direction of I.

What is ∆V across the unspecified circuit element? Does the potential increase or decrease when traveling through this element in the direction assigned to I?

1. ∆V increases by 2V in the direction of I.

2. ∆V decreases by 2V in the direction of I.

3. ∆V increases by 10V in the direction of I.

4. ∆V decreases by 10V in the direction of I.

5. ∆V increases by 26V in the direction of I.

9 V

Assume that the voltage of the battery

is 9 V and that the three resistors are

identical. What is the potential

difference across each resistor?

1) 12 V

2) zero

3) 3 V

4) 4 V

5) you need to know the actual value of R

Since the resistors are all equal,

the voltage will drop evenly

across the 3 resistors, with 1/3 of

9 V across each one. So we get a

3 V drop across each. 9 V

1) 12 V

2) zero

3) 3 V

4) 4 V

5) you need to know the actual value of R

Follow-up: What would be the potential difference if R= 1 Ω, 2 Ω, 3 Ω

Assume that the voltage of the battery

is 9 V and that the three resistors are

identical. What is the potential

difference across each resistor?

12 V

R1= 4 Ω R2= 2 Ω

In the circuit below, what is the

voltage across R1?

1) 12 V

2) zero

3) 6 V

4) 8 V

5) 4 V

12 V

R1= 4 Ω R2= 2 Ω

In the circuit below, what is the

voltage across R1?

1) 12 V

2) zero

3) 6 V

4) 8 V

5) 4 V

The voltage drop across R1 has

to be twice as big as the drop

across R2. This means that V1 =

8 V and V2 = 4 V. Or else you

could find the current I = V/R =

(12 V)/(6 Ω) = 2 A, then use

Ohm’s Law to get voltages.

Follow-up: What happens if the voltage is doubled?

In the circuit below, what is the

current through R1?

1) 10 A

2) zero

3) 5 A

4) 2 A

5) 7 A

10 V

R1= 5 Ω

R2= 2 Ω

The voltage is the same (10 V) across each

resistor because they are in parallel. Thus,

we can use Ohm’s Law, V1 = I1 R1 to find the

current I1 = 2 A.

In the circuit below, what is the

current through R1?

10 V

R1= 5 Ω

R2= 2 Ω

1) 10 A

2) zero

3) 5 A

4) 2 A

5) 7 A

Follow-up: What is the total current through the battery?

1) increases

2) remains the same

3) decreases

4) drops to zero

Points P and Q are connected to a

battery of fixed voltage. As more

resistors R are added to the parallel

circuit, what happens to the total

current in the circuit?

1) increases

2) remains the same

3) decreases

4) drops to zero

As we add parallel resistors, the overall

resistance of the circuit drops. Since V =

IR, and V is held constant by the battery,

when resistance decreases, the current

must increase.

Points P and Q are connected to a

battery of fixed voltage. As more

resistors R are added to the parallel

circuit, what happens to the total

current in the circuit?

Follow-up: What happens to the current through each resistor?

What is the potential at points a to e ?

1. Va = 0V, Vb = −4V, Vc = −10V, Vd = −12V, Ve = −20V

2. Va = −20V, Vb = −16V, Vc = −10V, Vd = −8V, Ve = 0V

3. Va = −4V, Vb = −6V, Vc = −2V, Vd = −8V, Ve = 0V

4. Va = 4V, Vb = 6V, Vc = 2V, Vd = 8V, Ve = 0V

5. Va = 20V, Vb = 16V, Vc = 10V, Vd = 8V, Ve = 0V

What is the potential at points a to e ?

1. Va = 0V, Vb = −4V, Vc = −10V, Vd = −12V, Ve = −20V

2. Va = −20V, Vb = −16V, Vc = −10V, Vd = −8V, Ve = 0V

3. Va = −4V, Vb = −6V, Vc = −2V, Vd = −8V, Ve = 0V

4. Va = 4V, Vb = 6V, Vc = 2V, Vd = 8V, Ve = 0V

5. Va = 20V, Vb = 16V, Vc = 10V, Vd = 8V, Ve = 0V

Which resistor has the

greatest current going

through it? Assume that all

the resistors are equal.

V

R1

R2

R3

R5

R4

1) R1

2) both R1 and R2 equally

3) R3 and R4

4) R5

5) all the same

V

R1

R2

R3

R5

R4

1) R1

2) both R1 and R2 equally

3) R3 and R4

4) R5

5) all the same

Which resistor has the

greatest current going

through it? Assume that all

the resistors are equal.

Follow-up: Which one has the smallest voltage drop?

The same current must flow through left and right combinations of resistors. On the LEFT, the current splits equally, so I1 = I2. On

the RIGHT, more current will go through R5 than R3 + R4

since the branch containing R5 has less resistance.

V

R1

R2

R3

R5

R4

Let’s compute the current in the previous question

I =V

ReqReq =

R1R2

R1 + R2+

(R3 + R4)R5

R3 + R4 + R5Ans:

Current flows through a

lightbulb. If a wire is now

connected across the bulb,

what happens?

1) all the current continues to flow through the bulb

2) half the current flows through the wire, the other half continues through the bulb

3) all the current flows through the wire

4) none of the above

The current divides based on the

ratio of the resistances. If one of the

resistances is zero, then ALL of the

current will flow through that path.

Current flows through a

lightbulb. If a wire is now

connected across the bulb,

what happens?

1) all the current continues to flow through the bulb

2) half the current flows through the wire, the other half continues through the bulb

3) all the current flows through the wire

4) none of the above

Follow-up: Doesn’t the wire have SOME resistance?

1) increase

2) decrease

3) stay the same

What happens to the voltage across

the resistor R1 when the switch is

closed? The voltage will:

V

R1

R3

R2

S

1) increase

2) decrease

3) stay the same

What happens to the voltage across

the resistor R1 when the switch is

closed? The voltage will:

With the switch closed, the addition of

R2 to R3 decreases the equivalent

resistance, so the current from the

battery increases. This will cause an

increase in the voltage across R1 .

V

R1

R3

R2

S

Follow-up: What happens to the current through R3?

Example: A a problem where Kirchoff’s rules are needed (neither parallel nor series)The ammeter in the figure reads 3.0 A. Find I1, I2 and E.

Ans: I1 = 1.0 A, I2 = 2.0 A, E = 15 V

Two lightbulbs operate at 120 V, but one

has a power rating of 25 W while the other

has a power rating of 100 W. Which one

has the greater resistance?

1) the 25 W bulb

2) the 100 W bulb

3) both have the same

4) this has nothing to do with resistance

Since P = V2 / R the bulb with the lower

power rating has to have the higher

resistance.

Two lightbulbs operate at 120 V, but one

has a power rating of 25 W while the other

has a power rating of 100 W. Which one

has the greater resistance?

1) the 25 W bulb

2) the 100 W bulb

3) both have the same

4) this has nothing to do with resistance

Follow-up: Which one carries the greater current?

Rank in order, from largest to

smallest, the powers Pa to Pd

dissipated in resistors a to d.

1. Pb > Pa = Pc = Pd

2. Pb = Pc > Pa > Pc

3. Pb = Pd > Pa > Pc

4. Pb > Pc > Pa > Pd

5. Pb > Pd > Pa > Pc

Rank in order, from largest to

smallest, the powers Pa to Pd

dissipated in resistors a to d.

1. Pb > Pa = Pc = Pd

2. Pb = Pc > Pa > Pc

3. Pb = Pd > Pa > Pc

4. Pb > Pc > Pa > Pd

5. Pb > Pd > Pa > Pc

V2/R (2V)2/R= 4 V2/R V2/(2R)= (1/2)V2/R V2/(R/2)= 2 V2/R

Rank in order, from

brightest to dimmest, the

identical bulbs A to D.

1. A = B = C = D

2. A > B > C = D

3. A > C > B > D

4. A > C = D > B

5. C = D > B > A

Rank in order, from

brightest to dimmest, the

identical bulbs A to D.

1. A = B = C = D

2. A > B > C = D

3. A > C > B > D

4. A > C = D > B

5. C = D > B > A

The current through A is the sum of those through B and C/D. So A is largest. The potential across B is the sum of the potentials across C and D separately, so the current through B is larger than the current through C. Finally, clearly the same current goes through C and D.

Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will:

1) glow brighter than before

2) glow just the same as before

3) glow dimmer than before

4) go out completely

5) explode

Since bulb B is bypassed by the wire,

the total resistance of the circuit

decreases. This means that the current

through bulb A increases.

Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will:

1) glow brighter than before

2) glow just the same as before

3) glow dimmer than before

4) go out completely

5) explode

Follow-up: What happens to bulb B?

1) circuit 1

2) circuit 2

3) both the same

4) it depends on R

The lightbulbs in the circuit below are

identical with the same resistance R.

Which circuit produces more light?

(brightness ⇐⇒ power)

1) circuit 1

2) circuit 2

3) both the same

4) it depends on R

The lightbulbs in the circuit below are

identical with the same resistance R.

Which circuit produces more light?

(brightness ⇐⇒ power)

In #1, the bulbs are in parallel,

lowering the total resistance of the

circuit. Thus, circuit #1 will draw

a higher current, which leads to

more light, because P = I V.

Example: A power line has 0.20 ohms resistance per kilometer of length. It carries 300 A of current. Find:(a) the voltage across 1.0 km of the wire, and(b) the power dissipated in each kilometer of the wire

Ans: (a) 60 V, (b) 18 kW

The time constant for the discharge of this capacitor is

1. 1 s.

2. 2 s.

3. 4 s.

4. 5 s.

5. The capacitor doesn’t discharge because the resistors cancel each other.

The time constant for the discharge of this capacitor is

1. 1 s.

2. 2 s.

3. 4 s.

4. 5 s.

5. The capacitor doesn’t discharge because the resistors cancel each other.

Two resistors in series

Equivalent resistanceis R=2Ω+2Ω =4Ω

Time constant τ = RC = 4 × 1 s = 4 s

In all of the circuits below the switch is initially open and the capacitor discharged.

Rank order the diagrams, form long to short, by the time it takes to charge the capacitor once the switch is closed.

1. b = c = d > a

2. d > b = c > a

3. b = c > a = d

4. a = d > b = c

5. b > a = c = d

S

2R

(b)

V C

(c)

R

2CV

S

C

S(a)

V

R

S

R

C

(d)

2V

In all of the circuits below the switch is initially open and the capacitor discharged.

Rank order the diagrams, form long to short, by the time it takes to charge the capacitor once the switch is closed.

1. b = c = d > a

2. d > b = c > a

3. b = c > a = d

4. a = d > b = c

5. b > a = c = d

The time constant is RC, independent of V

S

2R

(b)

V C

(c)

R

2CV

S

C

S(a)

V

R

S

R

C

(d)

2V

A battery, a capacitor, and a resistor are connected in series. Which of the following affect(s) the maximum charge stored on the capacitor?

1. the emf E of the battery

2. the capacitance C of the capacitor

3. the resistance R of the resistor

4. both E and C

5. all three of E, C, and R

A battery, a capacitor, and a resistor are connected in series. Which of the following affect(s) the maximum charge stored on the capacitor?

1. the emf E of the battery

2. the capacitance C of the capacitor

3. the resistance R of the resistor

4. both E and C

5. all three of E, C, and R

A battery, a capacitor, and a resistor are connected in series. Which of the following affect(s) the rate at which the capacitor charges?

1. the emf E of the battery

2. the capacitance C of the capacitor

3. the resistance R of the resistor

4. both C and R

5. all three of E, C, and R

A battery, a capacitor, and a resistor are connected in series. Which of the following affect(s) the rate at which the capacitor charges?

1. the emf E of the battery

2. the capacitance C of the capacitor

3. the resistance R of the resistor

4. both C and R

5. all three of E, C, and R

The rate at which the capacitor charges isprecisely the current through the resistor,I=dQ/dt

And

so the maximum rate is E/R but at later timesthe rate also depends on C (from the time constantRC in the exponent)

I =

E

Re!t/RC

Two resistors and an uncharged capacitor are connected to each other and then to a battery to form the circuit shown below.

Long after this is done, the currents I1 and I2 through the resistors R1 and R2, respectively, and the potential V across the capacitor are

1. I1 ≠ 0, I2 = 0, V = 0

2. I1 = 0, I2 = 0, V = 0

3. I1 ≠ 0, I2 = 0, V ≠ 0

4. I1 = 0, I2 ≠ 0, V ≠ 0

5. I1 ≠ 0, I2 ≠ 0, V ≠ 0

+– R2 C

R1

Two resistors and an uncharged capacitor are connected to each other and then to a battery to form the circuit shown below.

Long after this is done, the currents I1 and I2 through the resistors R1 and R2, respectively, and the potential V across the capacitor are

1. I1 ≠ 0, I2 = 0, V = 0

2. I1 = 0, I2 = 0, V = 0

3. I1 ≠ 0, I2 = 0, V ≠ 0

4. I1 = 0, I2 ≠ 0, V ≠ 0

5. I1 ≠ 0, I2 ≠ 0, V ≠ 0

+– R2 C

R1

Once the capacitor is charged no more current flows through it. The two resistors form a series circuit with the battery, so the current through them is non-zero. Also, the capacitor is charged, so V ≠ 0

Same situation: Two resistors and an uncharged capacitor are connected to each other and then to a battery to form the circuit shown below.

Long after this is done, the currents I1 and I2 through the resistors R1 and R2, respectively, and the potential V across the capacitor satisfy

1. I2 = V / R2

2. I1 = I2

3. I1 =E / (R1+R2)

4. all of the above

5. no more than two of the above

+– R2 C

R1

Same situation: Two resistors and an uncharged capacitor are connected to each other and then to a battery to form the circuit shown below.

Long after this is done, the currents I1 and I2 through the resistors R1 and R2, respectively, and the potential V across the capacitor satisfy

1. I2 = V / R2

2. I1 = I2

3. I1 =E / (R1+R2)

4. all of the above

5. no more than two of the above

+– R2 C

R1

Example: (a) What is the charge in the capacitor a long time after this circuit has been connected?(b) What is the current through the battery when this is initially set up, if the capacitor is initially discharged? (“initially set up” means... imagine having a switch connected to the battery, with the switch initially open)

Ans: (a) C E/2, (b) E/R

+–

R

C

R

more slides

1) increases

2) decreases

3) stays the same

V

R1

R3 R4

R2

S

What happens to the voltage

across the resistor R4 when the

switch is closed?

V

R1

R3 R4

R2

S

A B

C

We just saw that closing the switch causes an increase in the voltage across R1 (which is VAB). The

voltage of the battery is constant, so if VAB increases, then VBC must

decrease!

What happens to the voltage

across the resistor R4 when the

switch is closed?

1) increases

2) decreases

3) stays the same

Follow-up: What happens to the current through R4?

Two space heaters in your living

room are operated at 120 V. Heater 1

has twice the resistance of heater 2.

Which one will give off more heat?

1) heater 1

2) heater 2

3) both equally

Using P = V2 / R, the heater with the smaller resistance

will have the larger power output. Thus, heater 2 will

give off more heat.

Two space heaters in your living

room are operated at 120 V. Heater 1

has twice the resistance of heater 2.

Which one will give off more heat?

1) heater 1

2) heater 2

3) both equally

Follow-up: Which one carries the greater current?

1) twice as much

2) the same

3) 1/2 as much

4) 1/4 as much

5) 4 times as much

10 V

The three lightbulbs in the circuit all have the

same resistance of 1 Ω . By how much is the

brightness of bulb B greater or smaller than

the brightness of bulb A? (brightness ⇐⇒

power)

1) twice as much

2) the same

3) 1/2 as much

4) 1/4 as much

5) 4 times as much

10 V

We can use P = V2/R to compare the power:

PA = (VA)2/RA = (10 V) 2/1 Ω = 100 W

PB = (VB)2/RB = (5 V) 2/1 Ω = 25 W

The three light bulbs in the circuit all have the

same resistance of 1 Ω . By how much is the

brightness of bulb B greater or smaller than

the brightness of bulb A? (brightness ⇐⇒

power)

Follow-up: What is the total current in the circuit?

In the circuit shown, the two bulbs A and B are identical. Compared to bulb A,

1. bulb B glows more brightly

2. bulb B glows less brightly

3. bulb B glows just as brightly

4. answer depends on whether the mobile charges in the wires are positively or negatively charged

In the circuit shown, the two bulbs A and B are identical. Compared to bulb A,

1. bulb B glows more brightly

2. bulb B glows less brightly

3. bulb B glows just as brightly

4. answer depends on whether the mobile charges in the wires are positively or negatively charged

Bulbs are in series, hence same current I through them.

Brightness is related to power dissipated P = I2R; bulbs are identical, so R is the same, and since I is also the same, they dissipate same power, hence samebrightness.

In the circuit shown in (a), the two bulbs A and B are identical. Bulb B is removed and the circuit is completed as shown in (b). Compared to the brightness of bulb A in (a), bulb A in (b) is

1. brighter

2. less bright

3. just as bright

4. any of the above, depending on the rated wattage of the bulb

In the circuit shown in (a), the two bulbs A and B are identical. Bulb B is removed and the circuit is completed as shown in (b). Compared to the brightness of bulb A in (a), bulb A in (b) is

1. brighter

2. less bright

3. just as bright

4. any of the above, depending on the rated wattage of the bulb