have you tried substitute ohm's law into your KCL?
KCL states :current at a junction is zero
ohm states: I=E/R
something like I1=E/(1/R1 + 1/R2)-1
etc etc...does it helps?
Mistake in your second law,
This is not a complicated circuit, just use two variables for current if u want.
Replace i3 as (i1-i2)
See the circuit as two individual boxes, running your finger from the top left hand corner and try to apply the rule again.
Hz only means flow of electron, 1 Hz means 1 time per second to and fro,
can't see any relation with the output voltage. but it does effect higher current when the Hz is higher on a capacitive load, and lesser current on an inductive load
what do you know about copper?
it is conductive and non-magnetic material.
if it has no magnetic attraction , then what is the force that acts on the magnet ?
this is when the LENZ's law comes in,
for your case , magnetic fields dropped through the copper loop (conductor) and induces...
Force on a (current carrying conductor) this is the wire in your case, lying at right angle to a magnetic field can be given as:
F=BIl
B=flux density, magnetic field in your case
I= current
l=length of the wire
there is current in the loop that arrive at the same node, but the current is negligible. meaning not worth mentioning.
if you take one part of the loop at the same node to connect after the resistor near B, thereby creating another node, current that enters that series of 3 x 2Ω will meet at...
ya, key word here is net force, two forces accelerating at 6.1m/s/s, one of which was given, just get the N and magnitude correct for the other.
if you are the marker for my exam, i think i'll be gone! amended the units for acceleration...hahaha
this answer is correct,
next hint,
half of 1.5kg is 0.75 kg
if center is 1.4m, you will need another 1.4m to make this statement true, 2.8m
add 0.4kg to 1.5kg, which make this broom now 1.9kg
with these 3 values, you can easily work the answer as 1.1m
1) if R=1ohm
R eq= 2.5ohms
relation: (2R+R)x(2R+R) / (2R+R) + (2R+R) + R
2) inverse add last two R, answer will series add with the R in the line, u will eventually see only two R in parallel, then inverse add these last two to get your answer.
total mass 1.5kg
center of mass at 1.4m
imagine the broom is a 2.8m rod instead.
if you add 0.4kg to one side, what is the distance that you need to move from the center, in order to balance the rod.
a logical answer will be <1.4m
Thanks, you are great help, finally solved, valuable piece of info for charge re-distribution
Ca=8cb/12....eqn 1
Ca=18+6Cb/14...eqn 2
eqn1=eqn2
therefore, Cb = 5.4μF
therefore, 120 Ca = 80 x 5.4
Ca = 3.6μF
thanks, i am getting some headway now.
(Qa) = (Ca) x 120
=120μC if assume (Qa) is 1μF
120μC = (Cb) x 80
(Cb) = 1.5
thus, if (Ca)=1, (Cb)=1.5
the ratio is correct now, next hint please, on getting the correct answer.
Now, with the voltage across (Ca) increased to 140V after putting the 3μF in...
1. Homework Statement
d.c supply 200V
capacitors A and B connected in series
p.d across A is 120V
p.d. across A, increased to 140V when a 3μF capacitor is connected in parallel with B.
attached diagram for better understanding.
Find capacitance of A and B.
2. Homework Equations...
More current will flow to the branch with less resistance in a parallel circuit.
Total Current 2A
simplify the parallel branch as 3ohms vs 6 ohms, as (2+1) in series and (5+1) in series.
voltage across the main branch = 2A x 2ohms = 4V
because total resistance of the parallel branch is 2ohms...
ok, if supply is 200V, A is 3μF and B is 5μF connected in series.
total capacitance will be 1.8μF
total charge =1.8 X 200 =360μC
Voltage across A = Q/C(A)
= 360μC / 3μF
= 120V
Voltage across B = Q/C(B)
= 360μC / 5μF
= 72V
how do i apply this logic to the above if both A and B are unknown...
Homework Statement
d.c supply 200V
capacitors A and B connected in series
p.d across A is 120V
p.d. across A, increased to 140V when a 3μF capacitor is connected in parallel with B.
Find capacitance of A and B.
Homework Equations
The only hint is the voltage drop (20V) across B when 3μF...