Physics Urgent Help Needed!!

[XPFMember]

Assalamoalaikum!!
Can somebody please answer this question..it's urgent!!
A resistor is connected in series with a parallel combination of a bulb and variable resistor.
(a) Describe how the brightness of the light changes as the resistance of the variable resistor is increased from zero.
(b) Explain why it is possible to reduce the bulb current to zero by adjusting the variable resistor.

well for (a) part i think it will increase because increasing the resistance will increase the voltage across the var. res. and bulb (as they are in parallel) so causing the brightness to increase!

and (b) part i have no idea!

i'm nt sure of my first answer too...so i'll be glad if anyone can help!

JazakAllah Khair!

 

[larina]

Walaikumusalam.
well, for the first one (a), as it is connected in parallel, the voltage is the same throughout the circuit.
As the resistance increases, the current decreases, so less current will flow through the buld, and therefore it's brightness decreases.

for (b), as increasing the resistance in the variable resister, the current decreases.
the resistance in the variable resistor changes by change in the length..the more the length, the higher the reisistance.
the shorter the length, the less the resistance.
so shifting the knob of the variable resister to the end (where the length is the greatest) so the resistance will be greatest there, and therefore no current will flow.

 

[XPFMember]

so that means the brightness depends on the current..not the voltage?

 

[larina]

yes, always.

 

[sakibfaiyaz]

Brightness depends on power dissipated; ALWAYS!!

 

[beacon_of_light]

Assalamoalaikum!!
Can somebody please answer this question..it's urgent!!
A resistor is connected in series with a parallel combination of a bulb and variable resistor.
(a) Describe how the brightness of the light changes as the resistance of the variable resistor is increased from zero.

JazakAllah Khair!

When you increase the resistance of the variable resistor, voltage around the variable resistor increases and voltage around the "parallel combination" of the bulbs decreases. Therefore, power dissipated in the lamps reduces i.e. P = V^2 / R ...
As V decreases, power dissipated as light energy decreases . Hence, brightness reduces !

 

[larina]

^doesn't the brightness depend on the current???? :/
but thats what my teacher told me.

 

[beacon_of_light]

^doesn't the brightness depend on the current???? :/
but thats what my teacher told me.

Yup it depends on current as well ... but the idea is again the same i.e. low power dissipation ... P= I^2 x R
When circuit current decreases due to increase in the resistance , power dissipation also reduces in the parallel combination of bulbs but increases around the variable resistor due to heating effects ...