Why does the glow of lamps becomes weaker when heavy current appliance is switched on?

@pr159260-4NR0Kn • Oct 27, 2024

Oct 27, 2024

1.4K

I am completely Newbie in this electrical field. Sorry if i asked very basic questions here.

why does the glow of lamps becomes weaker when heavy current appliance is switched on? i have got answer in some site as that as heavy current appliance is switched on. total resistance decreases and current increase because of that. this in turn cause the greater voltage drop in line. How there will be voltage drop which causes the lamp glow to dim? how can this happen ? Help me in this doubt?

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lal

@lal-R60Xjx • Feb 25, 2013

Let us consider two cases- when the lamp alone is operational and when the lamp is operating with a heavy load in parallel.

First case: Say the lamp is a 60W incandescent bulb and phase voltage at this condition is 230V. The current drawn by the lamp is around 0.27A. Even if normal house wiring is done with copper wires, it do posses a little resistance. A low current like what the lamp consumes doesn't actually make a significant voltage drop (drop in this case will be way too less than a volt). So the terminal voltage can be considered to be 230V itself neglecting the very small drop.

Second case: Switching on a heavy load in parallel with the lamp. A heavy load draws large current from the circuit. A one kW heater will consume 4.4A at 230V. This current now will produce significant voltage drop which can be even a few volts in some cases. So the terminal voltage now decreases to (230 - voltage drop). Since the lamp is in parallel with this heavy load, this decrease in terminal voltage affects the lamp too which is the reason for that slight dimming.
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pr159260

@pr159260-4NR0Kn • Feb 25, 2013

ok sir thanks for your explanation. I got some idea from your explanation. I would like to relate with the formula..can you please tel me whether am i right?

R = V/I.. here now from your explanation that is To parallel of lamp load another load heater is added. Now as the heater load added,total resistance will be less than before adding heater. As per the R = V/I.. As Resistance decreased, I increased and voltage decreased. Am i right? is that correct to relate that with formula?
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lal

@lal-R60Xjx • Feb 26, 2013

Yeah that is it. Exactly. When loads work in parallel the effective impedance is actually decreasing owing to more current flow. If we consider a whole house as single unit of load to the power system, more appliences we switch on will cause the resistance of the house(as a single unit) to reduce which increases the current flow and thus increasing the voltage drop.
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pr159260

@pr159260-4NR0Kn • Feb 26, 2013

okay sir.I am now clear with this sir. thanks for your explanation. 😀
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Charul Sharma

@charul-FNfRfw • Jul 31, 2018

Sir if we cut a heater wire in two equal parts each part will have half resistance. Hence it will have four times power dissipation but this practice is not advisable. Why??

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Ramani Aswath

@ramani-VR4O43 • Aug 1, 2018

The current doubles making the elements melt.

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lal

@lal-R60Xjx • Nov 30, 2018

#-Link-Snipped-# What happens if a wire is loaded to 20A that is designed to carry only 10A? You do the math.

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