physconomics
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Battery has an EMF of 9V (negligible internal resistance), A and C each have a resistance of 6 ohms and B and D each have a resistance of 3 ohms.Name:  C741E2F5-7847-40F3-BE8E-DFCF9F7EC212.jpg.jpeg
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uberteknik
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(Original post by physconomics)
Battery has an EMF of 9V (negligible internal resistance), A and C each have a resistance of 6 ohms and B and D each have a resistance of 3 ohms.Name:  C741E2F5-7847-40F3-BE8E-DFCF9F7EC212.jpg.jpeg
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The resistances of lamps A and B are in series and form a potential divider. The same is true for lamps C and D.

Lamp E is placed across the mid points of these potential dividers.

Calculate both mid point p.d.'s and subtract to find the actual p.d. across lamp E.

Then rearrange P = V^2/R for each lamp to find the power dissipated in each and therefore relative brightness.
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physconomics
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(Original post by uberteknik)
The resistances of lamps A and B are in series and form a potential divider. The same is true for lamps C and D.

Lamp E is placed across the mid points of these potential dividers.

Calculate both mid point p.d.'s and subtract to find the actual p.d. across lamp E.

Then rearrange P = V^2/R for each lamp to find the power dissipated in each and therefore relative brightness.
A and B together in series have a resistance together of 9 ohms. C and D together also have a resistance of 9 ohms. Would this mean the pd across E would be zero? Thanks for your reply
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Sinnoh
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These questions are made easier if you remember Kirchhoff's 2nd law. Sum of potential differences across each circuit loop is equal to sum of electromotive force supplied.

(Original post by physconomics)
A and B together in series have a resistance together of 9 ohms. C and D together also have a resistance of 9 ohms. Would this mean the pd across E would be zero? Thanks for your reply
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uberteknik
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(Original post by physconomics)
A and B together in series have a resistance together of 9 ohms. C and D together also have a resistance of 9 ohms. Would this mean the pd across E would be zero? Thanks for your reply
The series resistance of both A,B and C,D is 9 ohms.

Now calculate the current through each series path. Use those currents to find the p.d. across the resistors A,B,C,D.

Because the ratios of the resistors A,B & C,D are identical, the p.d.'s wrt either of the battery terminals at the divider mid-points, will also be identical. The difference in potential across lamp E is therefore zero. No p.d. means no current through lamp E.
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