The supply voltage in a room is 120 v. The resistance of the lead wires is 6 . A 60 w bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 w heater is switched on in parallel to the bulb
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If you need only a fair estimate of the decrease in voltage across the bulb, we can proceed as follows.
The incremental current flow from the source due to the incoming heater will approximately be
i=240120=2Ai=240120=2A
This is responsible for the increase in voltage drop across the lead wire.
v=2×6=12V(res.1)(res.1)v=2×6=12V
If you require the exact value, we may calculate as follows.
We will assume that both the bulb and resistor are rated for 120 V. Their resistances will then respectively be 1440060=240Ω1440060=240Ω and 14400240=60Ω14400240=60Ω.
Let us calculate the voltage across the bulb in both cases.
Case 1: Only the bulb is ON
V1=120−1206+240×6=117.07VV1=120−1206+240×6=117.07V
Case 2: Both the bulb and heater are ON
The equivalent resistance across the bulb-heater combination is 240×60240+60=48Ω240×60240+60=48Ω
V2=120−1206+48×6=106.67VV2=120−1206+48×6=106.67V
We can see that the change of voltage across the bulb is
v=117.07−106.67=10.4V(res.2)(res.2)v=117.07−106.67=10.4V
If you prefer the voltage division technique over finding currents (which is more intuitive), we will proceed as follows.
v=120×(2406+240−486+48)=10.5V(res.3)(res.3)v=120×(2406+240−486+48)=10.5V
Res. 1, 2 and 3 are the required results. A variation in voltage of about 10.5 V (9%) is observed by the bulb.
Do note that both loads are highly resistive and we have chosen to safely ignore the effects of power factor.
The incremental current flow from the source due to the incoming heater will approximately be
i=240120=2Ai=240120=2A
This is responsible for the increase in voltage drop across the lead wire.
v=2×6=12V(res.1)(res.1)v=2×6=12V
If you require the exact value, we may calculate as follows.
We will assume that both the bulb and resistor are rated for 120 V. Their resistances will then respectively be 1440060=240Ω1440060=240Ω and 14400240=60Ω14400240=60Ω.
Let us calculate the voltage across the bulb in both cases.
Case 1: Only the bulb is ON
V1=120−1206+240×6=117.07VV1=120−1206+240×6=117.07V
Case 2: Both the bulb and heater are ON
The equivalent resistance across the bulb-heater combination is 240×60240+60=48Ω240×60240+60=48Ω
V2=120−1206+48×6=106.67VV2=120−1206+48×6=106.67V
We can see that the change of voltage across the bulb is
v=117.07−106.67=10.4V(res.2)(res.2)v=117.07−106.67=10.4V
If you prefer the voltage division technique over finding currents (which is more intuitive), we will proceed as follows.
v=120×(2406+240−486+48)=10.5V(res.3)(res.3)v=120×(2406+240−486+48)=10.5V
Res. 1, 2 and 3 are the required results. A variation in voltage of about 10.5 V (9%) is observed by the bulb.
Do note that both loads are highly resistive and we have chosen to safely ignore the effects of power factor.
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