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Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 25,002

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Electricity Problems

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In the given diagram R1 = 2 Ohms, R2 = 1 Ohm, R3 = 3 Ohms, R4 = 4 Ohms

The circular loop is conducting.

What is the equivalent resistance between A and C?[/*]
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What is the energy stored in the capacitor in MicroJoules?[/*]
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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

Bob

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Registered: 2010-06-20

Posts: 10,643

Re: Electricity Problems

hi

Let the potential difference between A and C be V.

Current can flow between these points by travelling through any of four resistances, and the total flow (I sub t) is just the sum of these four currents.

So

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 25,002

Website

Re: Electricity Problems

Is the Answer 12/25?

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Electricity Problems

I think it is. 

I'll have to look up some stuff for number 2.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 25,002

Website

Re: Electricity Problems

Someone solved it for me:

In steady state we know that there is no current flowing through the capacitor.

We mark the point of \(+8V\) as \(A\) and the point of \(+3V\) as \(B\) and the earthed point as \(C\)

Let the current flowing from \(A\) to \(B\) be \(i \)

Net drop through \(A\) to \(B\) should be \(5V\)

\(\Rightarrow 4i+i=5 V\)   (I have added drop across the two resistors equated it to 5)

\(\Rightarrow i=1 A\)

Also we can write

\({V}_{A}-{V}_{P}=4 V\)

\(\Rightarrow {V}_{P}=4 V\)

In the circuit \(PC\) no current is flowing hence the only drop occuring is due to the capacitor

Drop across \(PC= {V}_{P}-{V}_{C}=4 V \)

Drop across capacitor \(= 4 V\)

Energy in the capacitor \(= \frac{1}{2}C{V}^{2}=24 microJ\)

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.