Chapter 6: Direct Current Circuits

It flows in one direction

Direct current flows constantly in one direction, as from a battery. Reversing direction periodically is what alternating current, such as the mains, does; and transformers work only on AC.

Alternating current

Alternating current (AC), such as the mains supply, reverses its direction periodically, usually many times a second. Direct current is the opposite case: it flows constantly in one direction.

A constant voltage, steady in value and polarity

'DC voltage' is everyday shorthand for a constant voltage whose polarity never changes, though DC strictly means direct current. A voltage that reverses its polarity periodically is an AC voltage.

Its positive and negative terminals change places

Polarity is which terminal is positive. An AC voltage keeps exchanging its positive and negative terminals; a voltage that only rises and falls while staying positive is a varying DC voltage, not AC.

DC means direct current, and voltage is not current

Strictly, DC means direct current, so 'DC voltage' joins a current word to a voltage. Saying 'steady' twice would be redundant, not a contradiction; in practice the phrase means a constant voltage.

The 230 V mains at a wall socket

The mains is the manual's own example of AC: its current reverses direction many times a second. A radio's 13,8 V power supply runs from the mains but delivers DC, as cells and batteries do.

Electric potential difference

The manual follows industry in using 'voltage' for electric potential difference, measured in volts. Field strength is in volts per metre; charge per second is current, in amperes, and resistance is in ohms.

The same principles will apply to them

The manual says that when AC circuits come up, the same principles will apply. Kirchhoff's laws hold at every instant in an AC circuit too, so they are not limited to DC.

The total current into a junction equals the total out

At any junction the currents flowing in add up to the currents flowing out. None is used up there: charge cannot pile up or vanish at a junction, so less cannot flow out than flows in.

I = I1 + I2

The current arriving equals the currents leaving, so I = I1 + I2: the two parts add up to the whole. I = I1 − I2 treats one of the outgoing currents as flowing in.

1 A

Current in equals current out: 5 + 2 = 7 A arrives and 6 A leaves along the third wire, so 7 − 6 = 1 A leaves along the fourth. 13 A counts the 6 A as flowing in.

8 A

The supply current divides between the branches and joins up again, so it is their sum: 1,5 + 2,5 + 4 = 8 A; another branch would add more. 6,5 A leaves out the 1,5 A branch.

It is the sum of the resistor currents

By the current law the battery current divides between the branches, so it is all the branch currents added together; each resistor added draws more. The average would be the total divided by the number of branches.

Charge

Charge cannot pile up at a junction, so what flows in must flow out; the law rests on that (a current is moving charge), not on energy. If measured currents do not balance, a reading is wrong.

The flow in equals the two flows out added together

Water cannot pile up in the T, so the flow in equals the two outlet flows added, like current at a junction. None is used up in the connector, as no current is lost at a junction.

The total of the drops along any one path joining them

Add up the drops along any one route from one point to the other: every route gives the same total. Adding the totals of several routes would count the same voltage more than once.

4,5 V

The drops across series resistors add up to the battery voltage, so the other has 12 − 7,5 = 4,5 V. Series resistors share one current, not one voltage: 7,5 V on both would need a 15 V battery.

10 V

The drops across series resistors add up to the supply: 2 + 5 + 3 = 10 V. 4 V takes one drop as negative; 0 V is the closed-loop total, which counts the battery too.

3 V + 6 V = 9 V

By the voltage law the drops across series resistors add up to the battery voltage: 3 V + 6 V = 9 V. Adding the battery's 9 V as well counts the supply as another drop.

Exactly zero volts

Taking each voltage with its sign, the drops round any closed loop add up to zero; a battery passed from negative to positive counts as a negative drop. Leaving the battery out gives the battery voltage instead.

As a negative voltage drop

Going round, you pass through the battery from its negative to its positive terminal: a rise, not a drop. So it counts as a negative drop and the total comes to zero; counted as positive, nothing would cancel.

+6 V

Readings round a closed loop add up to zero: +4 − 10 + x = 0, so x = +6 V. −6 V gets the sign wrong, and +14 V adds the sizes, ignoring the signs.

Any path that returns to where it started

A closed loop is any route through a circuit that ends where it began, like the manual's trip round the outer loop from A back to A. 'Closed' here has nothing to do with a switch.

The voltage across a part, caused by the current through it

The manual defines a voltage drop as the voltage across a component such as a resistor, produced by the current through it (V = I × R). With no current there is no drop, so it is not a threshold.

Towards whichever end is positive

By the manual's convention the arrowhead points to the positive side. Current flows through a resistor from its positive end to its negative end, so the arrow points against the current, not with it.

The positive terminal

In the battery symbol the long line always marks the positive terminal and the short line the negative one; the manual adds a '+' beside it only to make that plain.

They say the same thing in different words

The manual calls them different but equivalent: the drops along any path between two points add up to the voltage between them, and round a closed loop they add up to zero. Both hold in any circuit.

A source of voltage to drive the current

A complete circuit needs a source of voltage to drive the current round it. The manual leaves the source out and assumes its positive terminal joins the left-hand wire. A switch is not needed for completeness.

It is the same as the battery voltage

R3 is connected straight across the battery, so the whole battery voltage is across it, whatever R1 and R2 do; those two share the same voltage between them. Only a resistor in series with others gets a part.

1070 Ω

In series the resistances simply add: 390 + 680 = 1070 Ω, more than either one. 248 Ω comes from the parallel formula, which always gives less than the smaller resistor.

2,1 kΩ

Series resistances add, so the other is what remains of the total: 3,3 − 1,2 = 2,1 kΩ. 4,5 kΩ adds the two given values, treating the total as if it were one of the resistors.

It increases

Series resistances add, so every resistor added raises the total, and a fixed supply then drives less current. The total falls only when resistors are added in parallel.

Both in series

In series the two resistances add, giving more than either alone, so the larger one on its own falls short. In parallel the result is lower than the smaller resistor, and a wire link is close to zero.

3,67 kΩ

Put all three in the same unit and add: 1000 + 2200 + 470 = 3670 Ω, or 3,67 kΩ. 473,2 Ω adds the kilohm figures as though they were ohms.

880 Ω

In series every resistance adds: 4 × 220 = 880 Ω. 55 Ω divides by four, the shortcut for equal resistors in parallel, which gives less than any one of them.

Each resistor carries the same current

A series circuit has only one path, so the same current flows through every resistor, whatever its value. The largest resistor does not carry less current; it takes the largest share of the voltage.

They all go out

A series string has only one path. An open circuit anywhere stops the current everywhere, so every lamp goes out. Only lamps wired in parallel carry on when one of them fails.

0,3 A

In series one current flows through both: 9,9 ÷ 33 = 0,3 A. 0,21 A puts the 33 Ω resistor's 9,9 V across the 47 Ω; 0,12 A divides it by the whole 80 Ω.

3 A

In series 1 + 2 = 3 Ω, so I = 9 ÷ 3 = 3 A. 13,5 A divides by the parallel combination, 0,67 Ω, as if the resistors were side by side.

0,28 A

In series 10 + 33 = 43 Ω, so I = 12 ÷ 43 = 0,279 A, nearest 0,28 A. 1,6 A divides by the parallel combination, 7,7 Ω, which applies only to resistors side by side.

3,2 mA

The series total is 1,5 + 2,2 = 3,7 kΩ, so I = 12 V ÷ 3700 Ω = 0,003 24 A = 3,24 mA, nearest 3,2 mA. 13 mA divides by the parallel combination, 892 Ω, instead.

2 mA

In series 2,7 + 1,8 = 4,5 kΩ, so I = 9 V ÷ 4500 Ω = 2 mA. 8,3 mA divides by the parallel combination, 1,08 kΩ, and 2 A forgets the kilohms.

40 V

The series resistance is 6 + 14 = 20 Ω, so V = I × R = 2 × 20 = 40 V. 8,4 V multiplies by the parallel combination, 4,2 Ω, which does not apply to resistors in series.

23,5 V

In series 120 + 820 = 940 Ω, so V = 0,025 A × 940 Ω = 23,5 V. 2,6 V uses the parallel combination, 105 Ω, and 23 500 V takes 25 mA as 25 A.

5 Ω

N equal resistors in parallel give R ÷ N: 10 ÷ 2 = 5 Ω. Adding the reciprocals, 1/10 + 1/10 = 0,2, is not the end: the reciprocal of that, 1 ÷ 0,2, is the answer.

17 Ω

For N equal resistors in parallel the result is R ÷ N: 51 ÷ 3 = 17 Ω. Adding the reciprocals gives 3 × 1/51 = 0,0588, which still has to be turned over: 1 ÷ 0,0588 = 17 Ω.

50 Ω

N equal resistors in parallel give R ÷ N: 1500 ÷ 30 = 50 Ω, the usual value for a dummy load. 500 Ω divides by 3 instead of 30, and 45 kΩ is the series total.

Most amateur transceivers are designed to work into it

A dummy load stands in for the antenna, so it must present the resistance the transmitter expects: 50 Ω for most amateur transceivers. Fifty 2,5 kΩ resistors happen to give 50 Ω; the count alone does not set it.

140 Ω

Seven equal resistors in parallel give R ÷ N = 1000 ÷ 7 = 142,9 Ω, nearest 140 Ω. 0,007 is the reciprocals added up, 7 × 1/1000, before the last step of turning that over.

210 Ω

1/R = 1/330 + 1/560 = 0,003 03 + 0,001 79 = 0,004 82, so R = 207,6 Ω, nearest 210 Ω. 165 Ω halves the smaller resistor, a shortcut that holds only for two equal ones.

720 Ω

1/R = 1/1200 + 1/1800 = 0,000 833 + 0,000 556 = 0,001 389, so R = 720 Ω, less than either. 600 Ω halves the smaller resistor, which is right only for two equal resistors.

1080 Ω

1/R = 1/270 − 1/360 = 0,003 70 − 0,002 78 = 0,000 926, so R = 1080 Ω; check: 360 Ω and 1080 Ω in parallel give 270 Ω. 630 Ω adds the two values, which only series does.

160 Ω

1/R = 1/270 + 1/560 + 1/1200 = 0,006 32, so R = 158,2 Ω, nearest 160 Ω. 90 Ω divides 270 Ω by three, as if all three were 270 Ω.

320 Ω

R1 and R2 in series make 400 Ω, in parallel with R3's 1600 Ω: 1/R = 1/400 + 1/1600, so R = 320 Ω. 93 Ω treats all three resistors as parallel.

1,3 kΩ

R1 and R2 in series make 1970 Ω, in parallel with 3900 Ω: 1/R = 1/1970 + 1/3900, so R = 1309 Ω, nearest 1,3 kΩ. 330 Ω treats all three resistors as parallel.

It stays practically the same

Each branch has the full battery voltage across it, and a good battery's voltage hardly changes when the lamp's branch opens, so the radio draws practically the same current. It does not take over the lamp's share.

It decreases

Each branch has the full supply voltage across it, so by I = V ÷ R its current falls when its resistance rises. It cannot remain the same, as the voltage across it does not change.

0,8 A

Both resistors have the same voltage across them and the same resistance, so by Ohm's Law each carries 0,8 A; the battery supplies 1,6 A in all. The 0,8 A is not shared between them.

1,4 A

Each resistor has the full 12 V: 12 ÷ 30 = 0,4 A and 12 ÷ 12 = 1 A, so the supply gives 1,4 A. 0,29 A divides by 42 Ω, the series total, which does not apply here.

The one with the lower resistance

With the same voltage across both, I = V ÷ R gives the 100 Ω resistor ten times the current of the 1 kΩ one. In series it is the larger resistor that takes more of the voltage.

Shunt

'Shunt' is another word for parallel; the manual notes it is common in high-power work. A cascade is a chain of stages one after another, nearer to series; stacks and loops are not connection types.

1,25

A reciprocal is 1 divided by the number: 1 ÷ 0,8 = 1,25. Changing the sign gives −0,8, the negative, which is a different idea; 0,64 is the square and 0,89 the square root.

4 W

Equal resistors in parallel share the power equally: each takes 100 ÷ 25 = 4 W and must be rated for at least that. Each taking the full 100 W would make 2500 W in all.

To test a transmitter without radiating a signal

A dummy load takes the antenna's place and turns the transmitter's output into heat, so nothing goes on the air and testing causes no interference. Matching an antenna is the job of a tuner.

So that it cannot radiate a signal like an antenna

Unshielded, the resistors and wires could act as an antenna and put a signal on the air; the closed tin stops that, and it also serves as a heat sink. Keeping fingers off is not the reason given.

5,6 V

Vout = Vin × R2 ÷ (R1 + R2) = 28 × 3 ÷ 15 = 5,6 V. 7 V divides by R1 alone instead of R1 + R2, and 22,4 V is the share across R1.

3,5 V

Vout = 15 × 1,2 ÷ (3,9 + 1,2) = 18 ÷ 5,1 = 3,53 V, nearest 3,5 V. 4,6 V divides by R1 alone instead of R1 + R2, and 11,5 V is the share across R1.

4,2 V

Each resistor's voltage is its share of the total: 28 × 1,2 ÷ (1,2 + 6,8) = 4,2 V across R1. 4,9 V divides by R2 alone instead of R1 + R2, and 23,8 V is the share across R2.

2,8 V

Vout = 9 × 680 ÷ (1500 + 680) = 6120 ÷ 2180 = 2,81 V, nearest 2,8 V. 4,1 V divides by R1 alone instead of R1 + R2, and 6,2 V is the share across R1.

2,35 kΩ

R2 must take a third of the 12 V, so it is a third of R1 + R2, which makes it half of R1: 2,35 kΩ. 1,57 kΩ is a third of R1 alone instead.

11 kΩ

R2 must take 2 V of the 12 V, a sixth, so R1 + R2 = 6 × 2,2 kΩ = 13,2 kΩ and R1 = 11 kΩ. 13,2 kΩ is the total of both resistors, not R1 alone.

7,2 V

R2 is 1 kΩ of the 3 kΩ total, so it takes a third of the input: Vin = 3 × 2,4 = 7,2 V. 4,8 V is only the voltage across R1; the input is across both resistors.

Half of it

Both equal resistors carry one current, so they have equal voltages, each half the input: Vout = Vin × R ÷ 2R = Vin ÷ 2. A quarter would need the input halved twice.

It doubles

The output is a fixed fraction of the input, Vout = Vin × R2 ÷ (R1 + R2), so doubling Vin doubles Vout. A divider does not hold its output steady when the input changes.

Almost no current is drawn by any load

The formula assumes negligible current is taken from the output, so the current through R1 also flows on through R2. Power ratings do not enter into it; any values of R1 and R2 will do.

It falls below 3,3 V

The load is in parallel with R2, so that part of the divider now has less resistance and takes a smaller share of the voltage. The formula holds only while the output current is negligible.

16

Vout = Vin × 10 ÷ (150 + 10) = Vin ÷ 16: it divides by (R1 + R2) ÷ R2. 15 is R1 ÷ R2, leaving R2 out of the total; 0,0625 is the fraction, not the divisor.