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# Calculating voltage in a potential divider circuit. Watch

1. Hello

I've been doing some revision of my own for OCR module P6, and I'd appreciate it if somebody could check the answer to just this one question so that I can check I'm not doing it all wrong! I can't find an answer booklet online, and there aren't any answers in the online textbook either

A 100Ω resistor and a 200Ω resistor are used in a potential divider circuit. When the supply voltage is 12V, what is the voltage across the 200Ω resistor (R2)?

I did this:

(200/300)*12=8 (the voltage out)
12-8=4
Therefore voltage across R2 is 4V.

Was I right or wrong? If I was wrong can you explain how I should have done it please? I've tried googling to check but my search remains fruitless!
Thankyou
2. (Original post by DayFlower)
Hello

I've been doing some revision of my own for OCR module P6, and I'd appreciate it if somebody could check the answer to just this one question so that I can check I'm not doing it all wrong! I can't find an answer booklet online, and there aren't any answers in the online textbook either

A 100Ω resistor and a 200Ω resistor are used in a potential divider circuit. When the supply voltage is 12V, what is the voltage across the 200Ω resistor (R2)?

I did this:

(200/300)*12=8 (the voltage out)
12-8=4
Therefore voltage across R2 is 4V.

Was I right or wrong? If I was wrong can you explain how I should have done it please? I've tried googling to check but my search remains fruitless!
Thankyou
It's easy enough to check this one.
The pds will be in the same ratio as the resistors, which is 200 to 100 or 2 to 1.
So they must be 8V and 4V as that's 2 to 1 and they must add up to 12V
And the larger voltage is across the larger resistance.
So it's 8V across the 200 and 4V across the 100

Your calc was correct and the 8V you found is the value you want as you used the 200 ohm value on the top of that fraction. No need to subtract it from 12.
3. Thank-you! I thought it looked a little odd, but I never really understood the involvement of ratios before - I just assumed that R2 would have the smaller value because it's below the output voltage in the circuit
Thanks again, it's become a lot clearer now

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