Recent content by bensm0

The calculation for the maximum bending moment is to be verified experimentally... The calculation for the maximum bending moment is to be verified experimentally using a strain gauge bonded to the outer surface of the beam, at the point where the maximum bending moment occurs. Derive an...

Am i right in saying i can use this formulae: - W/EI((x^3/12) - (xL^2/16)) + w/EI((Lx^3/12) - (x^4/24) - (xL^3/24)) please help. See attached question. Question aii. Thanks

ok so i think i may have the answer: poisson's ratio = - transverse strain/axial strain Transverse strain = (39.99 - 40)/40 = -0.25x10^-3 axial strain = -(-0.25x10^-3/0.3) = 833.333x10^-6 Axial stress = 833.333x10^-6 x 200x10^9 = 166.6666x10^6 Force = stress x area Force = 166.6666x10^6...

A mild stell bar 40 mm diameter and 100 mm long is subjected to a tensile force along its axis. Young's modulus of elasticity = 200 GN m -2. Poisson's ratio is 0.3. Calculate the force required to reduce the diameter to 39.99 mm It says use the x-y coordinate system?? This i don't...

Is dt going to be the reciprocal of the frequency? And will EMF be zero for max flux occurs when transformer is idle? I'm still unsure.

Could you explain that equation please?

A single phase transformer has the following rating: 120 kVA, 2000 V/100 V, 60 Hz with 1000 primary turns. Determine: (a) the secondary turns (b) the rated primary and secondary currents (c) the maximum flux (d) given a maximum flux density of 0.25 T, the cross-sectional area of the core...

Ive got it, phew. Yes i used V,out = (V,in * (R2*R3/R2+R3))/(R1+(R2*R3/R2+R3)) After solving an quadratic equation i found x=0.5. Phew! Thanks

Am i anywhere near?

Is it like this: V,out=V,in*R3/(R1+R2)?

so if i use the voltage divider rule: V,out=V,in*R2/(R1+R2) Could i say: R1=10000x Ω R2=10000(1-x) Ω R3=5000Ω and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??

so if i use the voltage divider rule: V,out=V,in*R2/(R1+R2) Could i say: R1=10000x Ω R2=10000(1-x) Ω R3=5000Ω and for the voltage divider rule R2 would be (5000 x 10000(1-x))/(5000 + 10000(1-x))??

If we call the resistor in series R1 and the resistor in parallel R2 and the 5kΩ load resistor R3. Can i have R1+R2=10kΩ? And because R2 has a voltage of 3V and therefore R1 has a voltage of 6V can i say R2/R1=3/6 and R2=0.5R1? I'm still a little unsure?

Do i not need to find the currents? I have 0.6 mA going through the 5kΩ load resistor. I'm not sure about that equation? Please help?

The circuit of (attached file) shows a 10 kΩ potentiometer with a 5 kΩ load. Determine the position of the slider on the ‘pot’ when the voltage across points ‘XX' is 3 V. I can't figure out how you would find the current first without knowing total resistance? Any help would be greatly...