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About voltage drop
About voltage drop
Every circuit has resistance which limits the flow of current, the voltage are affected by resistances of these cable.
When a circuit wired over a long distances or length, this can cause voltage drop. Practically before connecting any loads or electrical appliance on the supply isolator, always measure the voltage incoming at least 220 volts AC.
where
Vc - voltage drop on cable circuit Vo - voltage at output (or terminal voltage) Ib - current design (or operating current of load)mcb - miniature circuit breaker for protection against overcurrent.
Vc - voltage drop on cable circuit Vo - voltage at output (or terminal voltage) Ib - current design (or operating current of load)mcb - miniature circuit breaker for protection against overcurrent.
Cable voltage drop formula
Cable voltage drop formula
Voltage drop on circuit is denoted as Vc,
Vc = mV/A.pm x Ib x L ÷ 1000
where
mV/A.pm mili-volt ampere per meter (Refers to Table 4D1B)
Ib current design
L length of the cable installed
1000 mili-volt unit
Table 4D1B
Table 4D1B
Voltage drop properties as in IEE Wiring Regulation (16 Edition)
Example calculation
Example calculation
A load rated 3.5kW (240vAC) connected from DB to point A. The total distance of the circuit is 12 meter. The cable are using PVC and supported by conduit system.
a. find current design, Ib?
Solution: Ib = P ÷ V = (3.5 x 1000) ÷ 240v = 14.58 A
b. select the cable size suitable for this circuit ?
Solution: Cable size select shall be : 2.5 mm² PVC.
c. identity the cable voltage drop properties ?
Solution:
Refer to Table 4D1B the under Method 3 Column(3)
voltage drop properties for 2.5mm² is 18 mV/A.pm
d. find the voltage drop on circuit, Vc ?
Solution:
Vc = mV/A.pm x Ib x L ÷ 1000
= 18 mV/A.pm x 14.58 A x 12m ÷ 1000 = 3.149 volt
Conclusion :: The voltage drop on circuit is ACCEPTABLE since 4% voltage for 240vAC is 9.6 volt
2. A three phase motor rated 8HP (415v AC) with efficiency of 80% and power factor 0.8 connected from DB to point A. The total distance of the circuit is 8 meter. The cable are using PVC and supported by direct clip method.
a. find current design, Ib?
Solution: Ib = (HP x 746 ) ÷ (415V x 1.732 x eff x pf
= ( 8 x 746 ) ÷ ( 415v x 1.732 x 0.8 x 0.8 ) = 12.97 A
b. select the cable size suitable for this circuit ?
Solution: Cable size select shall be : 4.0 mm² PVC.
c. identity the cable voltage drop properties ?
Solution:
Refer to Table 4D1B the under Method 1 Column(7)
voltage drop properties for 4.0 mm² is 9.5 mV/A.pm
d. find the voltage drop on circuit, Vc ?
Solution:
Vc = mV/A.pm x Ib x L ÷ 1000
= 9.5 mV/A.pm x 12.97 A x 8m ÷ 1000 = 0.985 volt
Conclusion :: The voltage drop on circuit is ACCEPTABLE since 4% voltage for 415v AC is 16.6 volt
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