Parallel circuits have two or more pathways through which current can flow. The total value of the current depends on the parallel resistance offered against the current flow. Since current and resis-tance are indirectly related, the highest resisresis-tance will allow less current to flow, and the lowest resistance will allow more current to flow.
The total resistance of a set of parallel resistors is found in the following manner:
The value of the final resistance shows another characteristic of parallel resistances. The total resistance of a set of parallel resistors is always equal to or less than the value of the smallest resistor in the parallel set. When an extremely large resistor (500 meg-ohms) is
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in parallel with an extremely small resistor, the total resistance is equal to the smallest resistor. When a set of parallel resistors are all of equal resistance, we can divide the resistance value of one of them by the total number of resistors in parallel to find the resistance of the parallel system.
Example
Let’s do a problem where elements of both the series circuit and the parallel circuit are combined.
1.
1.1.
1.1. Find the total resistance for the circuit.Find the total resistance for the circuit.Find the total resistance for the circuit.Find the total resistance for the circuit.Find the total resistance for the circuit.
The first step to solve the problem is to find the equivalent resistance for the R2 – R3 – R4 parallel set of resistors:
1 1 1 1
1 1
10 1 8
1 4 1 25
4
1 2 3
R R R R
R
R R
t
t
t
t
= + +
= + +
=
=
Ω Ω Ω
Ω
Ω . CHAPTER 5
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Finally, we add the two parallel equivalent values to the values of R1, R5, and R8, yielding:
3. Find the voltage change between points B and C.
3. Find the voltage change between points B and C.
3. Find the voltage change between points B and C.
3. Find the voltage change between points B and C.
3. Find the voltage change between points B and C.
This may seem difficult at first, but remember Ohm’s Law. The current passing through R5 is 1A, and the resistance is 3Ω.
V = I R V = (1A)(3Ω) V = 3 V ELECTRIC CIRCUITS
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4.
4.4.
4.
4. F F F F Find the vind the vind the vind the vind the voltaoltaoltaoltaoltaggggge ce ce ce ce changhanghanghange betwhange betwe betwe betwe between points een points een points een points A and B.een points A and B.A and B.A and B.A and B.
Again, the problem seems more difficult than it really is—it’s another Ohm’s Law problem. The total current passing through the parallel resistors is 1A, and the equivalent resistance is 4Ω.
V IR
V A
V V
=
=
=
(1 )(4 ) 4
Ω
P
OWERPower in circuits is the rate at which electric energy is used. The power capability of the circuit elements is the product of the voltage and the current.
P = VI
Through substitution from Ohm’s Law, we see two other ways to calculate power:
P I R P V
=( )( )2 = R
2
and .
Power is an important quantity in circuits. The voltage source must have enough power to operate the devices in the circuit.
Furthermore, the devices in the circuitry will burn out and open if their power capacity is not large enough to perform work at the required rate. Remember, power is the rate at which work is done.
The power requirement for a circuit or a circuit element can be calculated if any two of the Ohm’s Law quantities are known.
Example
A typical power calculation could require you to find the power requirement for a resistor in a circuit.
Find the power dissipated by a 4Ω resistor that has a .05A current passing through it.
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P I R
P A
P
=
=
=
2
05 4 01
(. )( ) .
Ω Watt
Another type of power question you could be asked would require you to find the total resistance in a circuit, then solve to find either the total power requirement of the circuit or the power requirement for a single element in the circuit.
Example
Find the power requirement for the series–parallel circuit shown below:
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Solution
To find the total power for the circuit you first must find the total resistance. We find that the total resistance for the circuit is 9Ω. With this information we can now solve for the power required to run the circuit with P V
Another question might involve one of the parallel resistors in the diagram above. The current in the circuit would be required to solve the problem.
Example
Find the power capacity for R3 in the circuit above.
Solution
First we find the total current with Ohm’s law.
I V
The next step is to calculate the voltage change across the two parallel resistors (R2 and R3).
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With this information the actual current in R3 can be found by using Ohm’s Law.
Next we calculate the power.
P = (V)(I)
across the parallel resistors does not work because each resistor has a different current through it.
C
APACITORSCapacitors (previously mentioned) are parallel plate devices that are capable of storing electric charge and releasing (discharging) it as needed. Their effectiveness is enhanced by placing a non-conductive material, called a dielectric, between the two plates. The energy (in the form of the charge) stored by a capacitor is E=1CV
2
2. Capacitors placed in an electric circuit do not allow a steady current to flow. Instead, they build charge between their two plates over a period of time until the potential of the capacitor is almost equal to the applied voltage. When a capacitor is fully charged in a DC circuit, current cannot flow until the capacitor is discharged, because the capacitor acts like an open circuit element.
Capacitors used in an electric circuit can be in series or in
parallel. The total capacitance (Ct) for capacitors in a parallel circuit is found by adding the values of the capacitors in parallel with one another. You should note that this technique is the opposite of the technique we used to find the value of the resistance of parallel resistors.
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The six capacitors in the diagram have values of 2, 3, 4, 5, and 6 micro farads, respectively. The total capacitance is found by adding them together.
The total capacitance for capacitors in series with one another is found by the reciprocal method. (Again this is just the opposite of resistors in series).
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