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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?

sorry, didnt notice that your eq is also correct.

see my loop 2 above, also take note that your E=IR ok? solved?

Loop 1, -R1(I2)+E=0 Loop 2, -R2(I1-I2)+R1(I2)=0

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.

Read about lenz's law about opposing induced force. It satisfied the answer and in actual fact, they do not cancel out.

do you have the answer since it's a MCQ? Lenz's law?

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

magnetic flux have definite direction and they never cross path, always from North to South forming a complete closed loop. think about it.

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...

your penciled attempt work out to be 3μF, which is same as mine and i think it is correct.

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

unsolved, but thanks anyway.

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...

care to show how to solve these 2 unknown? :)

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...

really hope someone can help to explain. Am i posting in the correct section? I am new here.

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...

attached is the diagram i have prepared.

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...

Blur on forming a maths equation to solve this problem.

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...