Electrical theory 1.2b - how to calculate series circuits

In this section I'll explain how to calculate various electrical values for series and parallel circuits for DC (AC is different and of little interest to us).

Series circuit

Here's a series circuit made of these components. Note: there's only 1 current path and it flows through each component in the series circuit - as shown.

Components:

• batteries: connected + - + - + - + - + - + - means you add up all voltages to give 9V;
• LEDs have individual voltage drops according to the colour of light emitted - see LEDs. These LEDs are shown as conducting and emitting light;
• resistors have specific resistance as made. When current flows, it has a voltage drop. This resistor could be the resistance value of a relay coil;
• mechanical switch has closed resistance of 0 and open resistance of infinity;
• electronic switch like a transistor or MOSFET operate similar to a switch except the ON (closed) voltage drop is around 1.1V. Its OFF value of 10's volts (as rated).... so no current can flow;
• silicon diode is shown conducting so its voltage drop is 0.6 - 0.7V; if it was connected in reverse, its voltage drop is 100's volts (as rated).... so no current can flow.

Important points:

1. Light bulbs, resistors, relay coil & contact, and mechanical switches are calculated on resistance;
2. batteries, transformers, switching power supplies are calculated on voltage;
3. electronic components, eg LED are calculated primarily on voltage dropped across them;
4. the above is satisfactory for simple lighting circuits using DC.

HINT:
When doing series calculations, the supply voltage/ voltage drops are processed together and the resistance values are totalled. Then Ohm's Law is used to give the current.

Calculate Current value

Analogy: to work out the series equation, stand at the battery + terminal and write down what the likely current would see (note I'll call voltage drop =Vd ) all around its path back to negative terminal of battery:

battery voltage 9V sees (=) red LED Vd + blue LED Vd + resistor * current + switch resistance (closed) * current + closed electronic switch Vd + diode Vd

Putting in values     9V = 2V + 3.4V + 33* current + 0* current +1.5V + .6V 

so                            9V = (2+3.4+1.5+.6) V + (33 + 0)* current

so                               9V = 7.5V + 33* current

so                            33* current = 9 - 7.5 = 1.5V

using Ohm's Law, current = voltage / resistance

                                 current = 1.5/ 33 = 0.045A

It's this easy to calculate the current.

Hints for series circuits:

• add the resistance values to give a total value;
• add voltage drops and substract from supply voltage. This gives a net voltage value;
• use Ohm's Law to calculate current.

Calculate a specific resistance value

You can also work out resistance or voltage values by slightly changing the process. Let's assume the switch is faulty and it offers resistance in the closed state. We measure the current to be 0.03A ... what's the switch resistance (called X Ohms)?

Use the analogy again:

battery voltage 9V sees (=) red LED Vd + blue LED Vd + resistor * current + switch resistance (closed) * current + closed electronic switch Vd + diode Vd

Putting in values 9V = 2V + 3.4V + 33* current + X* current +1.5V + .6V

so                                   9V = (2+3.4+1.5+.6) V + (33 + X)* current

so                                   9V = 7.5V + (33+X)* current

so                                   (33+X)* current = 9 - 7.5 = 1.5V

using Ohm's Law, current = voltage / resistance, OR resistance = voltage/ current

current = 0.03A   so                   resistance = 1.5V/ 0.03A = 50 Ohms

so           33+X = 50, therefore X, switch closed resistance = 50-33 =17 Ohms

Calculate supply voltage value

You can also work out supply voltage value by slightly changing the process. Let's assume the switch is good and it offers zero resistance in the closed state. The current of 0.045A will burn out the LEDs and it must be reduced to 0.02A ... what's the new supply voltage (called Vx)?

Use the analogy again:

battery voltage Vx sees (=) red LED Vd + blue LED Vd + resistor * current + switch resistance (closed) * current + closed electronic switch Vd + diode Vd

Putting in values     Vx = 2V + 3.4V + 33* .02 + 0* .02 +1.5V + .6V

so                            Vx= (2+3.4+1.5+.6) V + (33 + 0)* .02

so                            Vx = 7.5V + (33)* .02 = 8.16V 

So reducing the supply voltage from 9V to 8.16V reduces the current from 0.045A to 0.02A.

We'll need the above processes when designing LED circuits. I'll fully explain each design step so you can follow the calculations.

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