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I’m not sure about this question is it less than 6V because X is 6volts and therefore the total would add up to more

Than 10V?

Than 10V?

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#3

Picture is kind of difficult to read

But the p.d across X is not 6V. The pd across X should be 4V

It is required that the p.d across Y is 6V

so we have Y/(Y+1.2)=6/10

using potential divider

hence solve for Y

But the p.d across X is not 6V. The pd across X should be 4V

It is required that the p.d across Y is 6V

so we have Y/(Y+1.2)=6/10

using potential divider

hence solve for Y

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#5

(Original post by

Is the resistance of Y 800 ohms?

**Bananasplitxxx**)Is the resistance of Y 800 ohms?

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It was this part I don’t understand, hopefully it’s a better picture

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#8

(Original post by

Is the resistance of Y 800 ohms?

**Bananasplitxxx**)Is the resistance of Y 800 ohms?

For the next question, think about what happens the p.d across Y & Z.

Since Y and Z are in parallel, the effective resistance of the parallel combination decreases (and is now less than that of either of the components Y and Z). Hence the p.d across Y and Z must be smaller than 6V.

For the last part, calculate the effective resistance of the parallel combination. Then use potential divider again. [Note: other methods are possible using Ohm's law but taking from the question, it is probably expected to use potential divider]

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#9

(Original post by

Hang on how do you get 1800 ohms?

**Bananasplitxxx**)Hang on how do you get 1800 ohms?

so we have Y/(Y+1.2)=6/10

using potential divider

hence solve for Y

Yes sorry, I didn't realise that you needed help only with the 2nd part. Picture was hard to read haha

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Sorry I’m still confused with part 1, what does the potential divider do?

Can’t i find the current through X using V=IR and then use that current to find the resistance of Y?

Can’t i find the current through X using V=IR and then use that current to find the resistance of Y?

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(Original post by

Sorry I’m still confused with part 1, what does the potential divider do?

Can’t i find the current through X using V=IR and then use that current to find the resistance of Y?

**Bananasplitxxx**)Sorry I’m still confused with part 1, what does the potential divider do?

Can’t i find the current through X using V=IR and then use that current to find the resistance of Y?

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#12

**Bananasplitxxx**)

Sorry I’m still confused with part 1, what does the potential divider do?

Can’t i find the current through X using V=IR and then use that current to find the resistance of Y?

Yes you can find current through X using V=IR and then use that current to find the resistance of Y

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#14

(Original post by

Can you please explain part 3 again

**Bananasplitxxx**)Can you please explain part 3 again

First find the resistance of the parallel combination using

1/R = 1/Y + 1/Z

where Y and Z are the resistances of Y and Z respectively

Hence from this find the effective resistance in the circuit which is X + the resistance of the parallel combination

Then use V=IR to find the p.d across X. Since the p.d across X + the p.d across parallel combination = 10

you can find the p.d across the parallel combination

the p.d across Y is the same as the p.d across Z since they are in parallel

[You can also use potential divider]

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#16

(Original post by

Is the resistance of Y still the same as the last part?

**Bananasplitxxx**)Is the resistance of Y still the same as the last part?

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#18

hence R = 1300 ohms

so V = 1300 / (1300+1800) x 10 = 4.2 V

the effective resistance of the parallel combination is 1300 ohms

so you can think of the parallel combination as a single resistor of 1300 ohms. the p.d across this single resistor is then the p.d across the parallel combination. this p.d is equal to the p.d across Y and equal to that across X. hence the p.d is calculated as above

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X

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