Ohm’s Law and Power Formulae
Series Circuit Characteristics
1. The current is the same everywhere in the circuit.
–
This means that wherever I try to measure the current, I will obtain the same
reading.
2. Each component has an individual Ohm's law Voltage Drop.
–
This means that I can calculate the voltage using Ohm's Law if I know the
current through the component and the resistance.
3. Kirchoff's Voltage Law (KVL) applies.
–
The sum of all the voltage sources is equal to the sum of all the voltage drops
–
VS = V1 + V2 + V3 + . . . + VN
4. The total resistance in the circuit is equal to the sum of the
individual resistances.
–
RT = R1 + R2 + R3 + . . . + RN
5. The sum of the power supplied by the source is equal to the
sum of the power dissipated in the components.
–
PT = P1 + P2 + P3 + . . . + PN
1
Sources Connected Series Aiding
Sources Connected Series Opposing
VT = VS1 – VS2 + VS3
2
Calculate the Voltage Drops
Characteristics of the Parallel Circuit
•
The voltage across each component (branch) is the same
everywhere in the circuit.
–
•
This means that wherever I try to measure the voltage, I will obtain the same
reading, and this is the supply voltage.
Each branch has an individual current path.
–
We may calculate the branch current using Ohm's Law if we know the
voltage across the component and the resistance.
•
Kirchoff's Current Law Applies. This means that the sum of
all the currents entering a node is equal to the sum of all the
currents leaving the node
•
The inverse of the total resistance in the circuit is equal to
inverse the sum of the inverse of the individual resistances.
IT = I1 + I2 + I3 + . . . + IN
1
1
1
1
1
=
+
+
+ ... +
RT
R1
R2
R3
RN
•
The sum of the power supplied by the source is equal to the
sum of the power dissipated in the components.
PT = P1 + P2 + P3 + . . . + PN
3
Kirchhoff’s Current Law
• Kirchhoff’s current Law
(KCL) can be stated as:
ΣI = 0
The algebraic sum of all
the currents entering and
leaving a node is equal to
zero
ΣIin = ΣIout
The algebraic sum of all
the currents entering a
node is equal to the
algebraic sum of all the
currents leaving a node
I = 400mA
Find IT
4