An electric bulb of 500 W at 100v is used in a circuit

[Shivang kohlii]

If the circuit has 200 V supply.
The resistance R that must be put in series with bulb so that it draws 500 w is?

2. relevant equations
P= v^2/r
I = v/r
3. My attempt at the solution

R= V^2/P r (bulb)= 10000/500 = 20 ohm
Now for 200 v supply ..
P= 500 W
V= 200V
Rnet = 20 + R

R+20= 80
R= 60 ohm
But that's not the answer.. why?

 

[NascentOxygen]

Do you require the bulb itself to dissipate 500W, or do you require the 200V source to supply 500W?

 

[Mark44]

If the circuit has 200 V supply.
The resistance R that must be put in series with bulb so that it draws 500 w is?

To be clear, does "it" refer to the resistor?

2. relevant equations
P= v^2/r
I = v/r
3. My attempt at the solution

R= V^2/P r (bulb)= 10000/500 = 20 ohm

Where did the 10000 figure come from?
Are you trying to calculate the resistance of the bulb without the extra resistor? If so, you have used the wrong value for the voltage.

Also, in a series circuit, the voltages across the two loads add to the voltage supplied to the circuit, and the current I is the same throughout the circuit.

Now for 200 v supply ..
P= 500 W
V= 200V
Rnet = 20 + R

R+20= 80
R= 60 ohm
But that's not the answer.. why?

 

[Shivang kohlii]

Do you require the bulb itself to dissipate 500W, or do you require the 200V source to supply 500W?

Bulb I guess

 

[Shivang kohlii]

To be clear, does "it" refer to the resistor?

Where did the 10000 figure come from?
Are you trying to calculate the resistance of the bulb without the extra resistor? If so, you have used the wrong value for the voltage.

Also, in a series circuit, the voltages across the two loads add to the voltage supplied to the circuit, and the current I is the same throughout the circuit.

1. Yeah it refers to bulb
2. 10000 is for V^2 where V = 100
3. So that means that what I have calculated is basically power across R + bulb instead of only bulb? Also if the resistance R was in parallel .. then my answer would have been correct??

 

[Mark44]

2. 10000 is for V^2 where V = 100

So again, where does 10000 come from? In the problem statement, you said that the circuit has a 200 V supply. Is the voltage drop across the bulb 100 V.?

3. So that means that what I have calculated is basically power across R + bulb instead of only bulb? Also if the resistance R was in parallel .. then my answer would have been correct??

Let's focus on your problem as stated, in which the bulb and the resistor are in series.

In a series circuit the voltage drops across the components add up to the supplied voltage, and the current (I) is the same throughout the circuit. In a parallel circuit, the voltage is the same throughout, and the currents through each component add up to the total current supplied.

 

[Shivang kohlii]

So again, where does 10000 come from? In the problem statement, you said that the circuit has a 200 V supply. Is the voltage drop across the bulb 100 V.?

Acc to question ifthe bulb is connected to 100V supply power dissipated by it will be of 500 W
I have used that info to find resistance of the bulb which will remain constant irrespective of circuit conditions
Now if the bulb is connected to 200 V supply and the bulb still dissipates 500W that means that the potential is being divided and bulb is not getting the pd of 200v , because of the resistor

 

[gneill]

Acc to question ifthe bulb is connected to 100V supply power dissipated by it will be of 500 W
I have used that info to find resistance of the bulb which will remain constant irrespective of circuit conditions
Now if the bulb is connected to 200 V supply and the bulb still dissipates 500W that means that the potential is being divided and bulb is not getting the pd of 200v , because of the resistor

So the resistor has to drop half of the 200 V supply. If you want an equal split of the total potential difference across the resistor and the bulb, what must be the relationship between their resistance values?