Tutorial 12 Solutions

(1)

Tutorial 12 Solutions

1. Two resistors, of 100 Ω and 200 Ω , are connected in series to a 6.0 V DC power supply. (a) Draw a circuit diagram.

6 V

100

200Ω

Ω

(b) What is the total resistance?

Series circuit, so just add the resistances together:

RT = R1+R2

= 100+200 = 300 Ω

(c) What current flows in the circuit?

Use Ohm’s law. Since we want to use the total voltage to find the current through the whole circuit, we need to use the total resistance:

V = IR

6.0 = I×300

I = 6.0

300

I = 0.02 A=20 mA

(d) What is the potential difference across each resistor?

Now that we know the current through each resistor (in a series circuit, the current is the same everywhere), we can use this current to find the potential difference (voltage) across each resistor, using Ohm’s law again.

For the 100 Ω resistor:

V = IR

= 0.02×100 = 2.0 V

(2)

V = IR

= 0.02×200 = 4.0 V

(e) What is the power lost in each resistor?

We can use P=VI to find the power loss. We need to use the voltage across each resistor

(not the total voltage) to find the power lost in each resistor. If we wanted to find the total power used by the circuit, we’d use the total voltage.

P = VI

= 2.0×0.02

= 0.04 W=40 mW

P = VI

= 4.0×0.02

= 0.08 W=80 mW

We can also use P= I2R:

P = I2R

= 0.022×100 = 0.04 W=40 mW For the 200 Ω resistor:

P = I2R

= 0.022×200 = 0.08 W=80 mW

(f) What is the total power used by the circuit?

We could repeat the above calculations, using P = VI and the total voltage, or P = IR

using the total resistance.

However, since the only power used is that lost in each resistor, we can just add the powers we found previously:

(3)

2. Two resistors, of 1.0 kΩ and 2.0 kΩ , are connected in parallel to a 12 V DC power supply. (a) Draw a circuit diagram.

12 V _{1.0 k}_Ω

2.0 kΩ

(b) What is the total resistance?

Parallel circuit, so we have to use the total resistance in parallel formula: 1 RT = 1 R1 + 1 R2 1 RT = 1 1000+ 1 2000 1 RT = 0.001+0.0005 1 RT = 0.0015 RT = _0.00151 RT = 667 Ω

(c) What current flows in the circuit?

Use Ohm’s law. Since we want to use the total voltage to find the current through the whole circuit, we need to use the total resistance:

V = IR

12 = I×667

I = 12

667

I = 0.018 A=18 mA

(d) What current flows through each resistor?

Since the resistors are in parallel, they both have 12 V across them. So we just use Ohm’s law:

For the 1.0 kΩ resistor:

V = IR

12 = I×1000

I = 12

1000

(4)

For the 2.0 kΩ resistor: V = IR 12 = I×2000 I = 12 2000 I = 0.006 A= 6 mA

Note that these two currents add up to give the total current we found in the previous part of the question.

(e) What is the power lost in each resistor? We can use either P=VI or P= I2R.

P = VI

= 12×0.012

= 0.144 W=144 mW

P = VI

= 12×0.006

= 0.072 W=72 mW

(f) What is the total power used by the circuit? Adding the power lost in each resistor:

Ptotal =0.144+0.072=0.216 W = 216 mW.

3. Five 1.0 kΩ resistors are connected to a 5.0 V power supply, with two in series connected in parallel with a third, and these three resistors then connected in series with another two.

1.0 kΩ 1.0 kΩ 1.0 kΩ 1.0 kΩ 1.0 kΩ 5 V

(5)

(a) What is the total resistance?

We need to do this in a few steps. We can only combine resistors that are in series or parallel; we can’t do a combination of both in one step. First, we find some resistors that are in series 1.0 kΩ 1.0 kΩ 1.0 kΩ 1.0 kΩ 1.0 kΩ 5 V

and add their resistances together:

RT = R1+R2

= 1000+1000

= 2000 Ω

We can now replace these two resistors in the circuit with their combined resistance: 1.0 kΩ 1.0 kΩ 1.0 kΩ 5 V Ω 2.0 k

Next, we find two resistors in parallel with each other: 1.0 kΩ 1.0 kΩ 1.0 kΩ 5 V Ω 2.0 k

(6)

and find their combined resistance: 1 RT = 1 R1 + 1 R2 1 RT = 1 2000+ 1 1000 1 RT = 0.0005+0.001 1 RT = 0.0015 RT = _0.00151 RT = 667 Ω

and replace them in the circuit with their combined resistance:

1.0 kΩ

1.0 kΩ 5 V

667 Ω

Now, all of the resistances are in series, and we can just add them together to find the total resistance:

RT = R1+R2+R3

= 1000+1000+667

= 2667 Ω

The total resistance is 2.7 kΩ

(b) ADVANCED What current flows through each resistor? First, we find the total current in the circuit:

V = IR

5.0 = I×2667

I = 5.0

2667

I = 0.0019 A= 1.9 mA

This is the current in the resistors at the top and bottom, which are in series with the rest of the circuit. All of the current must flow through these resistors.

(7)

The short way:

Note that the other resitors are in two parallel branches, one with a total resistance of 2.0 kΩ (2 resistors) and the other with a total resistance of 1.0 kΩ (1 resistor). Since these two branches are in parallel with each other, they must have the same potential difference across them. So, the one with twice the resistance of the other must have half the current (since V= IR).

So, the branch with 2.0 kΩ resistance must have 1

3 of the total current, so: I = 1₃×0.0019=0.00063 A = 0.63 mA.

The other branch has 2

3 of the total current: I = 23×0.0019=0.0013 A = 1.3 mA.

The long way:

The two resistors which have the current of 1.9 mA flowing through them have potential drops of

V = IR

= 0.0019×1000 = 1.9 V

across them.

The potential difference across the parallel portion of the circuit is:

V= 5.0−1.9−1.9=1.2 V

This is the voltage across both parallel branches. We can now use Ohm’s law to find the currents in each branch:

2.0 kΩ (two resistors) branch:

V = IR

1.2 = I×2000

I = 1.2

2000

I = 0.0006 A= 0.6 mA 1.0 kΩ (one resistor) branch:

V = IR

1.2 = I×1000

I = 1.2

1000

I = 0.0012 A= 1.2 mA

4. An electric heater uses 2,400 W of power.

(a) What current is needed if the supply voltage is 240 V?

P = VI

2400 = 240×I

I = 2400

240

(8)

(b) What current flows in the high voltage power lines (11,000 V) to provide this power? P = VI 2400 = 11000×I I = 2400 11000 I = 0.22 A

5. A voltage divider is one way to make a controllable power supply. A variable resistor, with a total resistance of 1.0 kΩ , is used to control the output voltage of the following ciruit:

1

V

_in

R

₂ out

The variable resistor is divided into two sections, R1and R2, which together make up the total

resistance, so R1+R2 =1.0 kΩ .

(a) If the input voltage of the circuit, Vin, is 12 V, what do we set R1and R2to if we want Vout

to be 4.5 V? Assume that there is no output current from the circuit.

Since there is no output current, there isn’t anything connected across the output termi-nals, so the total resistance in the circuit is 1.0 kΩ .

The total current in the circuit is

V = IR

12 = I×1000

I = 12

1000

I = 0.012 A=12 mA

The potential drop across R2is V = IR= 0.012R2. From the circuit diagram, we see that

this is the same as Vout. Since we want Vout =4.5,

V = IR

4.5 = 0.012×R2 R2 = _0.0124.5 R2 = 375 Ω R1= 1000−R2 =1000−375 =625 Ω .

(9)

(b) What happens if there is an output current?

If there is an output current, there must be an extra resistance connected in parallel with

R2(the extra resistance is the resistance of the circuit that the output voltage is connected

to). Unless we know what this extra resistance is, we can’t find the total resistance, and can’t find the current, so we won’t know what to set R1and R2to. If we do know the extra

resistance, then we could still go ahead and find R1and R2– the values would be different

to those we got in the first part of the question.

6. ADVANCED A 100 Ω resistor is connected to a 10 V power supply. (a) What current flows in the circuit?

V = IR

10 = I×100

I = 10

100

I = 0.1 A=100 mA

(b) If an ammeter with a resistance of 5 Ω is used to measure the current. What current will be measured?

To measure the current, we need to connect the ammeter in series, so the total resistance of the circuit becomes 100+5=105 Ω . The current is now

V = IR

10 = I×105

I = 10

105

I = 0.095 A=95 mA

(c) A voltmeter with a resistance of 10 kΩ is used to measure the potential difference across the resistor. What value will be measured? When would the measurement be inaccurate? The voltmeter is connected in parallel with the resistor. This won’t change the potential difference across the resistor, so we’ll still measure 10 V.

The voltmeter will, however, change the total resistance of the circuit (reducing it to about 99 Ω ), which will increase the total current. If there were other resistances in series with the one the voltage measurement is being made across, the current through them would increase as well. If the current through other resistances increases, the potential drops across them would also increase. This would reduce the voltage across the 100 Ω resistor by a small amount.

(10)

7. ADVANCED A 100 µF capacitor is charged to 12 V. (a) How much charge is stored on the capacitor?

Q = CV

= 100×10−6×12 = 1.2×10−4C

(b) The capacitor is connected to a 10 kΩ resistor. i. What is the time constant of the circuit?

τ = RC

= 10000×100×10−6 = 1.0 s

ii. What is the potential difference across the capacitor 2 seconds after the resistor is connected? V = V0exp(−t/τ) = 12×exp(−2/1) = 12×exp(−2) = 12×0.135 = 1.6 V