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HNC Electrical engineering

someone please help me with task C im stuck,

A circuit consists of two, 12V dc batteries connected in parallel, with their positive terminals connected
together. One source has an internal resistance of 2 and the other has an internal resistance of 4 Ω. Their
connections are brought out to two terminals, A & B. The terminals need to be connected to supply a light
source. The only available lamps are rated at 6 V, 1 W.


a) Draw a circuit diagram of the above arrangement.
b) Determine and draw the Thevenin equivalent circuit.
c) Calculate the currents in each branch of the network to ensure the light source is operating at its rated values.
d) Calculate the resistance of the light source.
e) Use a circuit simulator to confirm your findings above
Reply 1
Original post by Jeffers999
someone please help me with task C im stuck,

A circuit consists of two, 12V dc batteries connected in parallel, with their positive terminals connected
together. One source has an internal resistance of 2 and the other has an internal resistance of 4 Ω. Their
connections are brought out to two terminals, A & B. The terminals need to be connected to supply a light
source. The only available lamps are rated at 6 V, 1 W.


a) Draw a circuit diagram of the above arrangement.
b) Determine and draw the Thevenin equivalent circuit.
c) Calculate the currents in each branch of the network to ensure the light source is operating at its rated values.
d) Calculate the resistance of the light source.
e) Use a circuit simulator to confirm your findings above

Seems like you'll have to connect quite a lot of bulbs in parallel and have the internal resistances dropping 6V.

If you tried running a pair of bulbs in series to half the voltage on each bulb the internal resistance of the power supply would reduce the voltage to less than 6V per bulb...
Reply 2
Original post by Joinedup
Seems like you'll have to connect quite a lot of bulbs in parallel and have the internal resistances dropping 6V.

If you tried running a pair of bulbs in series to half the voltage on each bulb the internal resistance of the power supply would reduce the voltage to less than 6V per bulb...

The question only wants me to workcout the currents on each branch of the circuit aswell as then using that to work out if the light source will be operating and its rated values
Reply 3
Original post by Jeffers999
The question only wants me to workcout the currents on each branch of the circuit aswell as then using that to work out if the light source will be operating and its rated values

Is the 'light source' allowed to contain multiple bulbs?.. Seems oddly worded where it says the only lamps (plural) available are 1W 6V.. Like something has been missed out of the question.
Reply 4
Original post by Joinedup
Is the 'light source' allowed to contain multiple bulbs?.. Seems oddly worded where it says the only lamps (plural) available are 1W 6V.. Like something has been missed out of the question.

From what i can tell its just doing calcuoations for the one lamp being the light source
Reply 5
Only way I can make sense of it is for the voltage drop in each internal resistor to be 6V

That gives you the current from each battery individually using I=V/R

cos the batteries are in parallel the current to the 'light source' will be the sum of the current from the batteries.

The current for one lamp when operated at it's rated power is P/V and you could parallel up several lamps until the total current was equal to the current you calculated being supplied by the batteries. That would seem to satisfy the conditions in the question.
Original post by Joinedup
Is the 'light source' allowed to contain multiple bulbs?.. Seems oddly worded where it says the only lamps (plural) available are 1W 6V.. Like something has been missed out of the question.

I thought I was the only one that has "this feeling" - something was missing in the question.

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