# Series Circuits

## Voltage and current in series circuits

In series circuits the sum of voltages is equal to the applied voltage and can be expressed as

U = U_{1}+ U_{2}+ ... + U_{n}(1)

where

U = applied voltage from the battery or source (volts, V)

U_{1..n}= voltage over each resistor (volts, V)

From Ohms Law

U_{n}= I_{n}R_{n }(2)

where

I_{n}= current (amps, A)

R_{n}= resistance (ohms, Ω)

Since the currents in a series circuit are equal

I_{1}= I_{2}= .. = I_{n}(3)

- equation (1) can be modified to

R = R_{1}+ R_{2}+ .. + R_{n}(4)

where

R = total circuit resistance in the circuit (ohms, Ω)

### Example - Series Circuit

To a *12 V* car battery three resistors *1.25 Ω, 0.5 Ω *and *1.5 Ω* are connceted in series. The total resistance in the circuit can be calculated as

R = (1.0 Ω) + (0.5 Ω)+ (1.5 Ω)

= 3 Ω

The current in the series circuit can be calculated as

I = U / R

= 12 / 3

= 4 amps

The power dissipated in the circuit can be calculated

*P = R I ^{2 }*

* = (3 Ω) (4 amps) ^{2}*

* = 48 watts*