Objectives:
Lesson 5.
Review: Work, Potential difference, Current and Power: Ohm’s Law; Equivalent resistance - Resistors in series and
parallel
Procedure for computing the equivalent resistance series-parallel circuits
Procedure for computing powers used and currents in series-parallel circuits
Review:
1.
Electrical circuits: charges flow through electrical circuits instead of water.
a.
Work is the amount of energy spent moving some amount of charge from one point in the circuit to another point
in the circuit - measured in J.
b.
The potential difference is the amount of energy spent moving 1C of charge from one point in the circuit to
another point in the circuit - measured in J/C.
c.
The electrical current is the amount of charge flowing through one point in the circuit per unit of time - measured
in C/s=A.
d.
The power is the work done per a unit interval of time - measured in J/s=W; it is computed as P = V x I
e.
Ohm’s Law resistance is a property of a resistor; it can be measured by measuring the voltage across the resistor
and the current through the resistor R=V/I
f.
Equivalent resistance:
i.
Series
ii.
Parallel
Power, potential difference and currents in circuits in Series
Since the current through the resistors in series is the same asthe current Ieq through the equivalent resistor then we can compute the
potential difference across the individual resistors as
V1 = R1 x Ieq; ...
The power delivered to individual resistors in series as
P1 = Ieq2 x R1; ...
Example: Determine the current through each resistor, the potential difference across each resistor and the power dissipated by each
resistor in the circuit below.
Req = 2S + 4S + 6S = 12S
Note that the sum of the potential differences is equal to the potential difference across the battery. Now we can compute the power
dissipated by each of the resistors two ways as illustrated in our computation below.
When we compute the power delivered by the battery P = V x Ieq = 3V x 0.25A = 0.75W. Similarly, when we compute the power
dissipated by the equivalent resistor we find that
. We note that the power dissipated by the
equivalent resistor is the sum of the power dissipated by each of the resistors P1 + P2 + P3 = 0.125W + 0.25W + 0.375W = 0.75W.
This is an important result that you can use to check your computation. This is an important result because it shows that the
energy is conserved. The power that we deliver to the resistors is equal to the total dissipated by the resistors.
Power, potential difference and currents in circuits in parallel
Since the potential difference across the resistors in parallel is the same as the potential difference Veq across the equivalent resistor
then we can compute the current through the individual resistors as
; ...
The power delivered to the individual resistors in parallel
; ...
Example: Determine the current through each resistor, the potential difference across each resistor and the power dissipated by each
resistor in the circuit below.
In the case of resistors in parallel the potential difference across each of them is the same and it is equal to 6V. We can then compute
the current through each of the resistors
Note that the sum of the three currents through each resistors is equal to Ieq. Now we can compute the power delivered by the battery
and the power dissipated by each of the resistors:
We again note that the total power dissipated by all of the resistors equal to 18W + 9W + 6W = 33W. When we compute the power
delivered by the battery P = 5.5A x 6V = 33W we note that the two powers are equal as they should.
Circuits in series-parallel
As we said before the resistors can be connected in variety of combinations The diagram below show four examples how three
resistors can be connected. The first two (a and b) are simple because the rules for series and/or parallel circuits can be directly
applied. The two other circuits © and d) are more complex. For example, R3 is not parallel with R2 in the circuit c or R2 is not in series
with R1 in the circuit d. Why? On the other hand, R2 is in series with R1 in the circuit c and R3 is parallel with R2 in the circuit d. Why?
It is important to note that we can call resistors to be “resistors in parallel” if they connect to the same points in the circuit. It
is equally important to note that we can resistors to be “resistors in series” if there is no junction between them.
How to solve problems involving combination of resistors in series and in parallel.
1.
2.
3.
5.
Identify those resistors that can be called either in series or in parallel. Compute the equivalent resistance Req1.
Draw a new circuit with the equivalent resistor Req1.
In the new circuit, identify those resistors that can be called either in series or in parallel. Compute the equivalent
resistance Req2.
Draw a new circuit with the equivalent resistor Req2. Continue repeating steps 1. and 2. until there is only one resistor
in the circuit. This resistor has the resistance Reqtotal
Then compute Ieqtotal.
6.
Then work backward using the rules:
7.
resistors and the potential difference across each of the resistors.
Then compute the power dissipated by each of the resistors.
4.
; ... and V1 = R1 x Ieq; ... to compute the current through each of the
Example: Determine the current through each resistor, the potential difference across each resistor and the power dissipated by each
resistor in the circuit below.
Procedure:
Step 1. We note that the same current flows through the resistors 1S and 2S - the resistors are in series
Req1 = 1SS + 2S = 3S
Step 2. Redraw the circuit Fig 1
Step 3. The resistors 2S and 4S are resistors in series and we can compute the equivalent resistance
Req2 = 2S + 4S = 6S
Step 4. Redraw the circuit. Fig 2
Step 4 - repeating step 1 and 2. The resistors 3S and 6S in Fig 2. are resistors in parallel.
Step 5.
Step 6. In the last circuit the potential difference across the resistors 3S and 6S is the same 3V. Therefore the current through each of
these equivalent resistors is
We note that the sum of these currents is equal to 1.5A as it should be equal to Ieqtotal.
Going back to Fig. 1 we note that the resistors R3 = 2S and R4 = 4S are in series and therefore the current Ieq2 = 0.5A that flows
through Req2 is the same as the current that flows through each of these two resistors. We can then compute the potential difference
across these two resistors
Going back to the original diagram we note that the resistors R1 = 1S and R2 = 2S are in series and therefore the current Ieq1 = 1A
that flows through Req1 is the same as the current that flows through each of these two resistors. We can then compute the potential
difference across these two resistors
Note that in each case the sum of the potential differences is equal to 3V
Step 7. The power dissipated by each of the resistors:
The total power delivered by the battery is P = Ieqtotal x V= 1.5A x 3V = 4.5W.