Current DC circuits problem(s)
Hi, can someone help me with this problem? I dont understand the solutions... The answer is C ( 8 ohms btw)
Original post by Julia William
Hi, can someone help me with this problem? I dont understand the solutions... The answer is C ( 8 ohms btw)
This is the classic 'Wheatstone Bridge' circuit. i.e. 2 potential dividers connected in parallel to the same supply.
When the voltmeter reads 0V that must mean the pd's either side of the voltmeter are equal.
That means the the ratios of the resistances in both potential dividers must also be equal.
The ratio of the length of the resistance wire is all you need to set up a balance equation with that on one side and the ratio of the resistors (including unknown) on the other.
Then, rearrange the equation to make X the subject and hey presto.
Spoiler
(edited 10 years ago)
Thanks! Do you mind explaining this question too?
(question 107)
(question 107)
Original post by uberteknik
This is the classic 'Wheatstone Bridge' circuit. i.e. 2 potential dividers connected in parallel to the same supply.
When the voltmeter reads 0V that must mean the pd's either side of the voltmeter are equal.
That means the the ratios of the resistances in both potential dividers must also be equal.
The ratio of the length of the resistance wire is all you need to set up a balance equation with that on one side and the ratio of the resistors (including unknown) on the other.
Then, rearrange the equation to make X the subject and hey presto.
When the voltmeter reads 0V that must mean the pd's either side of the voltmeter are equal.
That means the the ratios of the resistances in both potential dividers must also be equal.
The ratio of the length of the resistance wire is all you need to set up a balance equation with that on one side and the ratio of the resistors (including unknown) on the other.
Then, rearrange the equation to make X the subject and hey presto.
Spoiler
Original post by Julia William
Thanks! Do you mind explaining this question too?
(question 107)
(question 107)
All of the lamps are connected in series, which means that when one of them 'blows', it will then be open-circuit. i.e. you can think of it's resistance as being infinite.
A good way of visualising what the answer will be is to consider the voltages and currents of a potential divider circuit:
Write down an expression for both the current I and the two voltages voltages V1 and V2 of a potential divider. And then investigate what happens to the current and voltage if one of the resistors is held constant (good lamp) and one of the resistors gets much bigger (bad lamp tending towards infinity).
Then apply that knowledge to the circuit in question.
What do you get?
Spoiler
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