Parallel Circuits
Resistors connected in parallel
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The sum of current in a parallel network can be expressed as
I = I1 + I2 + ... + In (1)
where
I = sum of currents (Amps)
The total resistance in a parallel circuit can be expressed as
1 / R = 1 / R1 + 1 / R2 + ... + 1 / Rn (2)
where
R = total resistance in the parallel network (Ohms, Ω)
Example - Parallel Network
The total resistance in a network with three resistors R1 = 10 Ω, R2 = 20 Ω and R3 = 30 Ω can be calculated as
1 / R = 1 / (10 Ω) + 1 / (20 Ω) + 1 / (30 Ω)
= 0.183 (1/Ω)
R = 1 / 0.183
= 5.46 (Ω)
Connected to a 12 V battery the sum of currents can be calculated
I = U / R
= (12 V) / (5.46 Ω)
= 2.2 Amps)
The current through each resistor can be calculated as
I1= U / R1
= (12 V) / (10 Ω)
= 1.2 (Amps)
I2= U / R2
= (12 V) / (20 Ω)
= 0.6 (Amps)
I3= U / R3
= (12 V) / (30 Ω)
= 0.4 (Amps)
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Related Topics
- Electrical - Amps and electrical wiring, AWG - wire gauge, electrical formulas, motors and units
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Related Documents
- Conductors - Conductors and conductivity
- Kirchhoff's Laws - Kirchhoff's current and voltage laws
- Potential Divider - Output voltage from a potential divider
- Resistance and Resistivity - Electrical resistance and resistivity
- Series Circuits - Voltage and current
