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# DC Electricity unit question Watch

1. Q and A.docx
I need help on the last 2 questions on the attachment. For part b iii, WHY can't you just do
V=IR
v=(0.56)(3.33)
V=1.86V

? I can't understand what they do in the MS attached.

With the final part wouldn't you want the resistance of the voltmeter to equal the internal resistance of the battery for maximum power? In the MS it says infinite resistance is ideal. But how? That would mean all current is blocked. Why would you want to block it?

THANKS SO MUCH
2. (Original post by krisshP)
Q and A.docx
I need help on the last 2 questions on the attachment. For part b iii, WHY can't you just do
V=IR
v=(0.56)(3.33)
V=1.86V

? I can't understand what they do in the MS attached.

With the final part wouldn't you want the resistance of the voltmeter to equal the internal resistance of the battery for maximum power? In the MS it says infinite resistance is ideal. But how? That would mean all current is blocked. Why would you want to block it?

THANKS SO MUCH
I can't open that file - try a screen grab (alt+prtscr) and paste it into paint or something.
3. (Original post by Joinedup)
I can't open that file - try a screen grab (alt+prtscr) and paste it into paint or something.
https://docs.google.com/file/d/0B-2a...it?usp=sharing

can you see it
4. yeah thanks.

---
For part b iii, WHY can't you just do
V=IR
v=(0.56)(3.33)
V=1.86V
You've got to recalculate the current after changing the circulit

it'd be
I=V/Rtotal
I=3/(3.33+0.4)
= 0.81 for the current in the new circuit

then use that current in V=IRexternal
=0.81*3.33

to get the pd between the terminals... or use voltage divider theory

------
we don't want to maximise the *power* being converted in the voltmeter. - you want it to give an accurate reading which is why you want a high resistance, this will reduce the effect on the pd you're attempting to measure. This is an equivalent statement to wanting a very low current passing through your voltmeter.
5. (Original post by Joinedup)
yeah thanks.

---

You've got to recalculate the current after changing the circulit

it'd be
I=V/Rtotal
I=3/(3.33+0.4)
= 0.81 for the current in the new circuit

then use that current in V=IRexternal
=0.81*3.33

to get the pd between the terminals... or use voltage divider theory

------
we don't want to maximise the *power* being converted in the voltmeter. - you want it to give an accurate reading which is why you want a high resistance, this will reduce the effect on the pd you're attempting to measure. This is an equivalent statement to wanting a very low current passing through your voltmeter.
thank you

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