physics

[ruby92]

i meant it is...typo  too much physics
and its Q6 not 5 part a

[astarmathsandphysics]

6a)Y is positively charged since drop must be repelled by Y and attracted by X to remain stationary. The electric force must balance gravity.
Is this the right question? V Simple

[ruby92]

yes thnkx 

[halosh92]

electricity question:
1.When a battery in a toy car delivers a current of 0.75A, the p.d across its terminals is 1.3 V. when the car's motion is reversed the current delivered by the battery and the p.d. across its terminals becomes 12A and o.80v respectively.
calculate for the battery
a)the emf
b)the internal resistance

2.physics oct nov 2005 q6 part A(explained in detail)

whats the answer to the first question though??

[astarmathsandphysics]

I am sure the elctricity question makes no sense.

[ruby92]

only out teacher seems to think it does and she gives such questions for the mock 
where we all end up

[ruby92]

Electricity:
Question:
1. State kirchoff's laws. use the laws to deduce the laws of Ia and Ib as shown in the figure.
The circuit is here:
http://tinypic.com/r/t5j3iu/5

2.the circuit is attached.We need to find I1 I2 and I3

could someone solve these immediately
i have an exam tommorrow 

[halosh92]

is this the answer for question 2?

I1= 0.44
I2=0.39
I3=0.834

??

[astarmathsandphysics]

When I get home

[astarmathsandphysics]

V2+V3=3
I2=2I3 whe I2 is current in 2 ohm resistor etc  by symmetry
2I2+3I2=3 so I2=3/5A IA=IB=3/5A

[ruby92]

V2+V3=3
I2=2I3 whe I2 is current in 2 ohm resistor etc  by symmetry
2I2+3I2=3 so I2=3/5A IA=IB=3/5A

this is the answer to the first question?

[astarmathsandphysics]

Net voltage around left loop is 1V
V=IR so 1=6I so I=1/6A
similarly I2=1/5 and I3=1/5+1/6 sine the currents are opposed

[ruby92]

this is what i did
in the circuit a third current -Ia-Ib is there
for the first loop going in an anti clockwise direction
2Ia+3(-Ia-Ib)-2=0
2Ia-3Ia-3Ib-2=0.
-1Ia-3Ib-2=0.....(1)
in the second loop.
going anticlockwise
-2Ib-3(-Ia-Ib)+2=0
-2Ib+3Ia+3Ib+2=0
1Ib+3Ia+2=0....(2)

solving simulatneously.....
Ib as -o.5
and Ia by substitution as -0.5

what exactly have i done wrong?

[Phosu]

its weird that i got this rite in the exam, but i cant solve it again,
haha heres how i did it.

Ia + Ib = Ic , heres the basic mistake which you did, which caused the whole thing to go rong.

Ia be a, and Ib be b

2-2a-3Ic=0
2-2b-3Ic=0

any 2 above can be used as equation 1

2-2a-3Ic=2-2b-3Ic

3Ic and 2 will get cancelled
you get an equation -2a+2b=0, this will be used as equation 2

now simultaneously solve
i will use loop a, and split Ic current

2-2a-3(a+b)=0
2-2a-3a+3b=0

1)2-5a+3b=0
2)-2a+2b=0

and voila!

[ruby92]

whts d final answer???
and wht have i done wrong??
Reply fast