Dc circuit

The lamp acts as a resistance reducing the voltage available for the motor . I think that the answer is 6.5 volts.
I am a retired accountant by the way!

 

5.5v. Everything is in series, so the voltage will be different to each item, but the current the same.

 

Drew, you are right - it's 5.5V across the motor.

That's the important term here - ACROSS. Voltages are always measured across two points.

The battery is at 0V at the -ve end (not really, but hey...*) and the +ve end is at 12V. When you measure across the battery, you therefore measure the difference between the two terminals = 12V.

In a series circuit like wot you have here, the voltage will drop across any resistance in that circuit - starting from 12V and ending at 0V - so the votages must add up to 12V in total.

If the lamp has 6.5V across it, then the motor will have the remaining = 5.5V

(By the way, that means that the lamp has a slightly greater resistance than the motor. The greater the resistance of an object in a circuit, the greater the voltage across it.)


(* 'convention')

 

5.5 volts.

 

And for the not-so-speedy ones...

 

And he is a spark!

 

Alright, so if the +ve to other side of motor is 5.5v, there is only 5.5v at that point. Doesn't that mean the lamp is supplied with only 5.5v(and therefore won't work, as won't the motor) ?

Mr. HandyAndy - Really

 

Not if it's a 6.5v motor and a 5.5v lamp

 

Think of the motor as a lamp.
If both lamps were 5watts, they would both be equal loads. Two in series would split the voltage in half. The bulbs would both receive 6v. If one bulb was a higher wattage, this proportion would shift. Think about your christmas lights. 20 x 12v lamps, all in series. 20x12=240v.

This is a classic "potential divider" circuit.

 

That's a good question.

If the motor was designed for 12V operation, then it would be running at around half-speed. Ditto for the lamp.

If both were designed for 6V, then both would be getting nigh-on what they needed and would be running at full chat.

 

Exactly. You'd be lucky to light a 12v lamp at all with 6.5v.

Mr. HandyAndy - Really

 

It would light up quite noticeably - ~half-brightness.

And the motor would run at around half-speed.

 

Just over quarter brightness assuming it is a basic incandescant - power = V^2/R

so, (6.5/12)*(6.5/12) = 0.29 or 29%

 

That's confused me!

Are you saying that if an incandescent lamp gets half the leccy power it'll glow at only 29% brightness? Or is it that the lamp is only getting 29% of the actual power? Or even that incandescent lamps aren't linear in operation?!

I can't say I've followed your equation - since we don't know the resistance of the lamp - but whichever way I look at this, the lamp is getting around half the power, not a third.

Eg, if we - for simplicity's sake - assume the circuit is drawing 1A, then the resistance of the lamp would be 6.5ohms (R=V/I = 6.5/1) and the motor 5.5ohms.

Therefore the power consumed by the lamp would be P=VI which is 6.5 * 1 = 6.5W. If the motor was by-passed, the lamp would get the full 12v across it = 12 * 1 = 12W.

 

The lamp will be getting HALF voltage. In your example you have assumed that when the motor is removed the current remains the same - it will not as about half the resistance has been removed and thus the current would increase

Consider the level this question is aimed at and it would be fair to assume that the lamp would be a filament bulb. The resistance of a bulb does increase as current flows however for the purpose of this, keep it constant.

If the bulb has a resistance of 10R (Ohms) and is designed to operate at 12v then the current would be 1.2A and power calculated bu either IV,, V^2/R or I^2xR all of which give 14.4W

If the voltage is cut to 6.5v then the current flow will reduce to 0.65A and the power to 4.225W or 29%

 

You are soooo right

Removing the motor would (roughly) halve the resistance so double the current - D'oh.

Blimey, I can't believe I had to work that out so many times

 

12v potential and you are given 6.5v potential difference across the bulb means a 5.5v drop over the motor. Realistically values would be different but Voltage is a potential difference between two points.

6.5 - 12 = 5.5v

Potential divider...

This is a prime up question to a potential divider later on, they will add resistance values to calculate voltages and current around a circuit.

http://www.bbc.co.uk/schools/gcsebitesize/design/electronics/calculationsrev2.shtml