Question related to circuits. help me
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# Question related to circuits. help me

[From: ] [author: ] [Date: 11-05-19] [Hit: ]
Since the 2nd equation is a smaller value ( R-40 is larger than R-100), hence voltage across heater drops & there is less output.-Imagine we are supplying the circuit with a voltage V. The current through the circuit will be V/R where R is the total resistance of the circuit. For a series circuit, the resistances add.......
A 100 W bulb is connected in series with a room heater . What will happen if the bulb is replaced by a 40W bulb?

(A) Heater output will increase

(B) Heater output will decrease

(C) Heater output will remain unchanged

(D) Bulb will not glow.

i searched on internet but i get mixed answer like some one says remain unchanged, one says it will decrease.
tell me with proper explaination

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Strangely enough, I would say (B) decreases.

This is because a 40W bulb has a higher resistance than a 100W bulb.
Remember that Watts = V^2 / R
So if we have a fixed voltage (power supply of 220Vac), then a lower wattage is produced when R is higher.

Here's what happens with the heater in series -
lets say 100W bulb resistance is R-100 & 40W is R-40 & heater is R-heater

Voltage across heater with 100w bulb is -- V x [ R-heater / ( R-heater + R-100 ) ]
and with 40W bulb -- V x [ R-heater / ( R-heater + R-40 ) ]

Since the 2nd equation is a smaller value ( R-40 is larger than R-100), hence voltage across heater drops & there is less output.

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Imagine we are supplying the circuit with a voltage V. The current through the circuit will be V/R where R is the total resistance of the circuit. For a series circuit, the resistances add. Say R1 is the resistance of the lamp and R2 is that of the heater. The current will be:

I = V/(R1+R2)

now the power output of the heater will be:

P heater = I^2*R2 = V^2*R2/(R1+R2)

so you see it is inverseley proportional to the resistance of the lamp R1. A lower wattage lamp has a higher ressitance. Here is why:

If a just a lamp of ressitance R is connected to a voltage source V, the power is V^2/R. So a lower powered lamp has a higher resistance for the same voltage.

So back to the power of the heater, if we reduce the power of the lamp, we increase the lamps resistance, we can see from the equation that this decreases the heaters power output. Another way to think of it is that we increase the total resistance of the cirucit, so less current must flow (and remember that since it is a sereis circuit the current through the lamp and the heater are the same), and the power in the heater is proportional to the square of the current. So reducing the current in the circuit will reduce the output of the heater.

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I would say that the heater output has a high chance of increasing, a lower watt bulb consumes less voltage... It would be more conclusive if the question gave the voltages... Since we are connected in series, the voltage is shared between the bulb and the heater, hence lower demand of the bulb causes a higher output in heater...

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It would be A, Heater output will be increased.
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