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Laser and Optoelectronics Engineering Department DC circuits analysis laboratory 2011- 2012
Experiment NO.
Series and parallel connection
Object
To study the properties of series and parallel connection.
Apparatus
- DC circuit training system
- Set of wires.
- DC Power supply
- Digital A.V.O. meter
Theory
1. The Series Circuit
A SERIES CIRCUIT or ‘‘series-connected circuit’’ is a circuit having JUST ONE CURRENT PATH. Thus, Fig.(1) is an example of a ‘‘series circuit’’ in which a battery of constant potential difference V volts, and three resistances, are all connected ‘‘in series.’’
There are two ways to connect the resistance
1. Resistance in series
When some conductors having resistance R 1 – R 2 – R 3 etc. are joined end – on
- end as in (fig 3-1 a), these resistances are connected in series.
Fig.(1)
Laser and Optoelectronics Engineering Department DC circuits analysis laboratory 2011- 2012
Since a series circuit has just one current path, it follows that all the components in a series circuit CARRY THE SAME CURRENT I , a fact evident from inspection of Fig.(1). The current I is assumed to be a flow of positive charge, and thus flows out of the positive terminal of the battery and around through the external circuit, reentering the battery at the negative terminal. This is indicated by the arrows in Fig.(1). In a series circuit, the TOTAL resistance, RT, that the battery sees is equal to the SUM of the individual resistances. Thus, in the particular case of Fig.(1) the battery sees a total resistance, RT = R1 + R2 + R3 , while in the general case of ‘‘n’’ resistances connected in series the battery sees a total resistance of : RT = R1 + R2 + R3 +…..Rn
By Ohm’s law, it follows that the current I in a series circuit is equal to
Resistance, on the other hand, consumes electrical energy, removing it from the circuit in the form of heat. Since resistance does not produce or generate electrical energy, it is a non-active or PASSIVE type of circuit element. The potential difference between the terminals of a resistor is called the VOLTAGE DROP across the resistor, and, is equal to the current I times the resistance R ; that is, the ‘‘ voltage drop ’’ across a resistance of R ohms carrying a current of I amperes is I R volts.
We have the important fact that:
In a series circuit, the applied voltage is equal to the sum of the voltage drops.
Laser and Optoelectronics Engineering Department DC circuits analysis laboratory 2011- 2012
The currents in the individual resistances are called the ‘‘branch currents,’’ and the battery current I is often called the ‘‘line current.’’ From inspection of Fig.(3) we see that, in a parallel circuit, the battery current I is equal to the sum of the branch currents.
If the battery voltage V is applied equally to all n resistances; that is, the same voltage V is applied to all the parallel branches. Hence, by Ohm’s law, the individual branch currents in Fig.(3) have the values:
Then, we have:
Now let RT be the total resistance as seen by the battery in Fig.(3). Then, by Ohm’s law, it has to be true that:
Since the left-hand sides of the last two equations are equal, the two right- hand sides are also equal. Setting the two right-hand sides equal, then canceling the Vs, gives
Laser and Optoelectronics Engineering Department DC circuits analysis laboratory 2011- 2012
Procedure
- Using the DC circuit trainer, connect the circuit Shown in Fig. (4), take
VT =10V, and R 1 =82Ω, R 2 = 100Ω and R 3 =150Ω.
- Measured the voltage and current of "R 1 , R 2 & R 3 ", then record it in table below
- By using ohm's law, Calculate the RT
- Disconnect the DC power supply, and then measured the equivalent resistance by using the AVO meter only.
- Using the DC circuit trainer, connect the circuit Shown in Fig.(5), and
take VT =10V, and R 1 =82Ω, R 2 = 100Ω and R 3 =150Ω.
- Measured the voltage and current of "R 1 , R 2 & R 3 ", then record it in table below
V(volt) VT^ =
I(mA) IT^ =
Fig.( 4 )
V
V
V
10 V
Laser and Optoelectronics Engineering Department DC circuits analysis laboratory 2011- 2012
- In Figure, the battery voltage is V = 65 volts, and the values of the resistances, in ohms, are 38, 17, and 27, as shown. Find: (a) Total resistance seen by the battery, (b) Current measured by the ammeters shown in the figure, (c) Power output of the battery, (d) Power input to each resistor.
