Example 1.1

In Figure 1 E1 = 15 volts, E2 = 10 volts, V1 = 1.0 volt, V2 = 1.5 volts, what is V3 the voltage across R3?


Figure 1. A series circuit, starting at the lower left corner a battery goes up with the positive side at the top. A resistor labeled R1 goes from the top of the battery to the right. The other end of the resistor connects to a second resistor labeled R2 which continues to the right. A third resistor connects to the right end of R2. The right end of the third resistor labeled R3 connects to the top, positive end, of a battery which goes down. The negative side of the battery on the right connects to the negative side of the battery on the left, completing the circuit. The battery on the left is labeled E1 = 15 volts. The battery on the right is labeled E2 = 10 volts.


Solution

E1 - V1 - V2 - V3 - E2 = 0

Solving for V3 gives

V3 = E1 - V1 - V2 - E2

V3 = 15 - 1.0 - 1.5 - 10 = 2.5 volts

You stand a better chance of getting the right answer if you don't take shortcuts but write it out this way.

BACK

Example 1.2

In Figure 1 if the current in R2 is 1.62 amps what is the current in; (a) the battery at E1, (b) R1, (c) R3, and (d) the battery at E2?

Solution

The current is the same in all elements. All four answers are 1.62 amps.

BACK











Example 1.3

In the circuit of Figure 2 the battery current I is 127 mA, I2 is 33 mA, and I3 is 45 mA. What is I1?


Figure 2. A parallel circuit, starting at the lower left corner a battery labeled E1 goes up with the positive side at the top. A resistor labeled R1 connects to the top of the battery. The other end of the resistor connects to the bottom of the battery. A second resistor labeled R2 has it's top end connected to the top of the battery. This resistor is drawn to the right of the first one rather than over it. The bottom of R2 connects to the bottom of the battery. A third resistor labeled R3 has one end connected to the top of the battery. The bottom end of this resistor connects to the bottom of the battery. This third resistor is drawn to the right of the second one.


Solution

I = I1 + I2 + I3

Solving for I1 gives

I1 = I - I2 - I3

I1 = 127 - 33 - 45 = 49 mA.

BACK

Example 1.4

In the circuit of Figure 2 the battery voltage is 22.5 volts. What is the voltage across; (a) R1, (b) R2, and (c) R3?

Solution

In a simple parallel circuit all voltages are the same. All three answers are 22.5 volts.

BACK











Example 1.5

A measurement on a 2 cell flashlight shows that the bulb current is 350 milliamps. What is the hot resistance of the bulb's filament? One cell produces 1.5 volts.

Solution

We use the form of Ohm's law that will give us resistance.

R = V / I

V = 2 x 1.5 volts = 3.0 volts

I = 0.350 amps (given)

R = 3.0 / 0.350 = 8.57 ohms

If you used a calculator you likely got 8.571429 and maybe even more digits. These extra digits are not significant because the data we started with doesn't have that many digits. In this series of articles we will round off all answers to 3 digits. Leading zeros such as in the number 0.000359 are not significant. That number has 3 significant digits not 6.

BACK

Example 1.6

You need a resistor that will drop 125 volts when the current is 26.6 milliamps. What is the value of this resistor in (a) ohms, and (b) kilohms?

Solution

You must work ohm's law in volts, ohms, and amps. But there is a shortcut. You can work it in volts, k ohms, and milliamps or even volts, Meg ohms and microamps. So let's work in k ohms and milliamps.

R = V / I

R in k ohms = V in volts / I in milliamps

R = 125 V / 26.6 mA = 4.699248 k ohms.

Round it off to 4.70 k ohms. The number of ohms will be a larger number than the number of k ohms. To convert to ohms multiply k ohms by 1000.

R in ohms = R in k ohms x 1000

R in ohms = 4.70 x 1000 = 4700 ohms.

BACK

Example 1.7

If you connect a 25 k ohm (same as kilohm) resistor across an 850 volt power supply, how much current will flow? Give your answer in m A (same as milliamps).

Solution

So let's work in k ohms and mA.

I in mA = V in volts / R in k ohms

I = 850 V / 25 k ohms = 34.0 mA

BACK








Example 1.8

Four resistors are connected in series. They are; 2.2 k ohm, 3.3 k ohm, 2.7 k ohm, and 2.4 k ohm. What is the equivalent resistance?

Solution

The one equivalent resistor that could take the place of all four is the sum of the individual resistors.

Req = R1 + R2 + R3 + R4

Req = 2.2 k + 3.3 k + 2.7 k + 2.4 k = 10.6 k ohms

Notice that the sum is larger than the largest resistor. This is a good logical check for when you are working problems that you don't know the answer to and the result of a wrong answer could be a butned out circuit instesd of a low grade.

BACK

Example 1.9

Four resistors are connected in parallel. They are; 2.2 k ohm, 3.3 k ohm, 2.7 k ohm, and 2.4 k ohm. What is the equivalent resistance?

Solution

The one equivalent resistor that could take the place of all four is one over the sum of the one overs individual resistors.

Req = 1 / (1/R1 + 1/R2 + 1/R3 + 1/R4)

Req = 1 /(1/2.2 k + 1/ 3.3 k + 1/2.7 k + 1/2.4 k)

Req = 1/(0.000454545 + 0.000303030 + 0.000370370 + 0.000416666)

Req = 1/0.00154461 = 0.6474 k ohms = 647 ohms.

BACK

Example 1.10

In Figure 1 if R1 = 22 ohms, R2 = 18 ohms, and R3 = 12 ohms, what is the current through R1?

Solution

In a series circuit the current is the same everywhere so to ask for the current in R1 is to ask for the current in any part of the circuit.

We don't need to be concerned with the voltage across individual resistors. We can find the total resistance and the total voltage. That's all we need to calculate the current.

We begin by writing the voltage sum equation.

E1 = V1 + V2 + V3 + E2

Solving this equation for V1 + V2 + V3 gives,

V1 + V2 + V3 = E1 - E2

The voltage across all three resistors is,

V1 + V2 + V3 = 15 - 10 = 5 volts

The sum of all three resistors is,

RT = R1 + R2 + R3 = 22 ohms + 18 ohms + 12 ohms = 52 ohms

So the current is the total voltage divided by the total resistance,

I = VT / RT = 5 volts / 52 ohms = 0.0962 A = 96.2 mA

BACK

Example 1.11

In Figure 2 if E1 = 6 volts, R1 = 22 ohms, R2 = 18 ohms, and R3 = 12 ohms, what is the battery current?

Solution

The battery current is equal to the battery voltage divided by the equivalent resistance connected across it.

Req = 1/(1/R1 + 1/R2 + 1/R3)

Req = 1/(1/22 + 1/18 + 1/12)

Req = 1/(0.04545 + 0.05555 + 0.08333)

Req = 1/0.18434

Req = 5.4246 ohms

I = V / R = 6 volts / 5.4246 ohms = 1.11 A

BACK

Example 1.12

In Figure 2 if the voltage across R1 is 77 volts, what is the voltage across R3?

Solution

Figure 2 is a simple parallel circuit. The voltage across any element in a parallel circuit is the same as the voltage across any other element. Therefore the answer is 77 volts. Don't you wish they were all that easy.

BACK








Example 1.13

In Figure 1 if the current in R3 is 28 m A, what is the current in (a) R2, (b) R1?

Solution

Figure 1 is a simple series circuit. In a series circuit the current through any element is the same as any other element. Therefore the answers are (a) 28 mA, and (b) 28 mA. Hmm, maybe they are all that easy.

BACK








Example 1-14

In Figure 1 the voltage across R1 is 1.9 volts, and the voltage across R2 is 0.8 volts. What is the voltage across R3?

Solution

The voltage sums are written as follows,

E1 = V1 + V2 + V3 + E2

Solving for V3 which is the voltage across R3 we have,

V3 = E1 - E2 - V1 - V2

V3 = 15 v - 10 v - 1.9 v - 0.8 v = 2.3 Volts.

BACK

Example 1-15

In Figure 2 The current in R1 is 1.3 Amps, the current in R2 is 2.5 Amps and the battery current is 10 Amps. What is the current in R3?

Solution

Letting the current in R1 be I1 and so on we have,

I = I1 + I2 + I3

Solving for I3 we have,

I3 = I - I1 - I2

I3 = 10 A - 1.3 A - 2.5 A = 6.2 A

BACK

Example 1-16

In Figure 1 the voltage drop across R1 is 1.25 volts, the drop across R2 is 1.50 volts, and the circuit current is 0.682 m A. What is the resistance of R3? Hint. This is what I call a two level problem. You have to make an initial calculation and then use the result of that calculation to calculate the final answer.

Solution

To calculate the resistance of R3 we need to know the current through that resistor and the voltage across that particular resistor, not the voltage across some other resistor. Because this is a series circuit we are given the current as 0.682 mA. We need to calculate the voltage using voltage sums.

E1 = V1 + V2 + V3 + E2

V3 = E1 - E2 - V1 - V2

V3 = 15 v - 10 v - 1.25 v - 1.50 v = 2.25 Volts

R3 = V3 / I = 2.25 v / .682 mA = 3.30 k ohms.

BACK

Example 1.17

In the circuit of Figure 2 the battery current is 0.675 amps, the current in R2 is 0.15 amps, and the current in R3 is 0.275 amps. The resistance of R3 is 200 ohms. What is the resistance of R1? This is a three level problem.

Solution

To calculate the resistance of R1 we need to know the current through R1 and the voltage across it. We are not given either one directly. We are given the current in R3 and its resistance. From this we can calculate the voltage across R3.

V3 = I3 x R3 = 0.275 A x 200 ohms = 55 Volts.

Because this is a parallel circuit the voltage across R3 is the same as all the other voltages so the voltage across R1 is 55 volts. Now for the current in R1.

I1 = I - I2 - I3 = 0.675 A - 0.15 A - 0.275 A = 0.250 A

The resistance of R1 is,

R1 = V1 / I1 = 55 v / 0.25 A = 220 ohms

BACK

Example 1.18

You want to connect a resistor substitution box across the 120 volt AC power line. Don't ask why, you just want to. The box uses 1 watt resistors. The values you can select are as follows

15 ohm, 22, 33, 47, 68, 100, 150, 220, 330, 470, 680, 1 k ohm, 1.5 k, 2.2 k, 3.3 k, 4.7 k, 6.8 k, 10 k, 15 k, 22 k, 33 k, 47 k, 68 k, 100 k, 150 k, 220 k, 330 k, 470 k, 680 k, 1 Meg ohm, 1.5 Meg, 2.2 Meg, 3.3 Meg, 4.7 Meg, 6.8 Meg, and 10 Meg ohm.

What is the lowest value you can set the box to without burning out a resistor (exceeding 1 watt)?

Solution

Let's take the V squared over R formula and solve it for R as follows.

P = V squared / R

R = V squared / P = (120 x 120) square volts / 1 W = 14400 ohms = 14.4 k ohms.

Don't ask me what a square volt is.

The two closest settings are 10 k ohms and 15 k ohms. If we set the resistance below 14.4 k ohms the power will exceed 1 watt. The setting to use is 15 k ohms. And that's my final answer.

BACK

Example 1.19

Express the following numbers in scientific notation.

(a) 5280
(b) 26.48
(c) 0.0000036
(d) 525.8
(e) 0.0162
(f) 74890000
(g) 0.000000125
(h) 152000
(i) 9730000000
(j) 0.0000000000022

Solution

(a) 5280 = 5.28 x 103 = 5.28e3
(b) 26.48 = 2.648 x 101 = 2.648e1
(c) 0.0000036 = 3.6 x 10-6 = 3.6e-6
(d) 525.8 = 5.258 x 102 = 5.258e2
(e) 0.0162 = 1.62 x 10-2 = 1.62e-2
(f) 74890000 = 7.489 x 107 = 7.489e7
(g) 0.000000125 1.25 x 10-7 = 1.25e-7
(h) 152000 = 1.52 x 105 = 1.52e5
(i) 9730000000 = 9.73 x 109 = 9.73e9
(j) 0.0000000000022 = 2.2 x 10-12 = 2.2e-12

BACK

Example 1.20

Fix each of these numbers to be correctly expressed in scientific notation.

(a) 47.0 x 103
(b) 0.025e-6
(c) 28.7 x 106
(d) 365 x 10-12
(e) 455e3

Solution

(a)

47.0 x 103
Smaller, Larger
4.70 x 104

(b)

0.025 x 10-6
Larger, Smaller
2.5 x 10-8

(c)

28.7 x 106
Smaller, Larger
2.87 x 107

(d)

365 x 10-12
Smaller, Larger
3.65 x 10-10

(e)

455 x 103
Smaller, Larger
4.55 x 105

BACK

Example 1.21

Perform the following calculations and express the results in correct form.

(a) 1/4.7 x 10-6

In scientific notation 1 = 1 x 100

Mantissa = 1/4.7 = 0.212766

Exponent = 0 -(-6) = 6

1/4.7 x 10-6 = 0.212766 x 106

0.212766 x 106
Larger, Smaller
2.12766 x 105

Rounding off

2.13 x 105

(b) 2.5 x 10-3/4.7 x 103

M = 2.5/4.7 = 0.5319149

E = -3 - (3) = -6

2.5 x 10-3/4.7 x 103 = 0.5319149 x 10-6

0.5319149 x 10-6
Larger, Smaller
5.319149 x 10-7

Rounding off

5.32 x 10-7

(c) 1.5 x 106 x 2.2 x 10-6

M = 1.5 x 2.2 = 3.3

E = 6 + (-6) = 0

1.5 x 106 x 2.2 x 10-6 = 3.3 x 100

This result can be expressed without writing the power of ten.

1.5 x 106 x 2.2 x 10-6 = 3.3

(d) 1.2 x 103 + 8.2 x 102

First we must make the powers of ten match.

1.2 x 103 + 0.82 x 103

1.2 + 0.82 = 2.02

Putting it back with its power of ten,

2.02 x 103

BACK