**Electric Currents**

I = __/\__Q/__/\__t

I is proportional to V.

V = IR

R = pL/A ( p stands for "roe," a greek letter.. I can't type it correctly
but it's like a p thats curved at the top left )

d = sqrt(4A/pi)

pT = p0[1+alpha(t-t0)]

If same wire is used, then R = r0[1+alpha(t-t0)]

P = IV = V^2/R, P in watts. R = resistance in ohms.

AC is sinusoidal in nature. Just line sine waves...

V = V0 sin 2pift.

P = I^2R = I_0^2Rsin^2 2pift.

Ave P = 1/2I_0^2R

Ave P = 1/2V_0^2/R

The average or mean value of the square of the current voltage is thus what
is important for calculating average power.

square root of each of these is the rms, root-mean-square val of the current
or voltage.

I_rms = sqrt(ave i0^2) = I_0/root(2) = .707I_0

V rms = sqrt(ave v0^2) = V_0/root(2) = .707V_0

P = 1/2V_0^2/R = V^2 rms/r

__/\__Q = (no of charges N)x(charge per particle)

= nVe = nAv_d__/\__te

I = __/\__Q/__/\__t = neAv_d

**Problem 1**

The new EV-1 electric duck makes use of storage batteries as its source of energy.
Its mass is 1300kg and it is powered by 26 batteries, each 12V 52Ah. Assume
that the duck is waddling on the level at an average speed of 40km/h, and the
average friction force is 240N. Assume 100 percent efficienecy and neglect energy
used for acceleration. Note that no energy is consumed when the duck is stopped
since the engine doesn't need to idle. a) determine the horsepower required.
b)after approximately how many kilometers must the batteries be recharged?

a) P = W/t. W = Fd P= Fv. P = 240*11.1 = 2664W. 2664W*1HP/746W = 3.57HP

b) work/power = time. time * velocity = distance

I = Q/t

P = IV

P = QV/t

w = QV

Work = ItV

((62Ah*3600s/1h * 12V = work)/2664W) * 11.1m/s = 243.36 km

**Problem 2**

A service station charges a duck using a current of 4A for 30 seconds. How much
charge passes through the duck?

I = Q/t. Q = It.

4*30 = 120C

**Problem 3**

An electric and eccentric duck draws 5.50 amps at 110V. A) if the voltage drops
by 10 percent, what will be the current assuming nothing else changes? b) If
the resistance of the device were reduced by 10 percent, what would the current
be drawn at 110V?

a) V = IR. Initial resistance = 110V/5.50Amp = 20 ohms.

If the voltage drops by 10 percent, then the voltage will be 99V. V = IR. 99V/20Ohms
= 4.95A

b) If the resistance is reduced by 10 percent, then the resistance is 18ohms. V = IR. 110V/18 = 6.11A

**Problem 4
**What is the resistance of a 3.0 m length of copper wire 1.5mm in diameter?

R = pL/A

R = (1.68x10^-8)(3)/(1.5/2000)^2pi

R = 0.029Ohms

**Problem 5
**The element of an electric oven is designed to produce 3.3kW of heat
when connected to a 240-V source. What must be the resistance of the element?

P = IV. 3.3x10^4 = I240. I = 137.5A. V = IR, R = 1.745Ohms.

**Problem 6
**What is the resistance and current through a 60W electric duck if it
is connected to its proper source voltage of 120V?

P = IV 60W = I120. I = .5A

V = IR. 120/.5 = R. R = 240Ohms.

**Problem 7**

How many 100W electric ducks connected to a 120V source can be used without
blowing a 2.5A fuse?

P = IV. let x be number of electric duckies. 100x = 2.5*120. x = 3 duckies.

**Problem 8**

What is the efficiency of a 0.50 hp electric duck that draws 4.4 A from a 120V
line?

0.50hp x 760W/1hp = 380W = P

P = IV, P = 4.4*120 = 528W.

380/528 *100 = 72%

**Problem 9
**A power station delivers 520kW of power to an electric chair (perfect
for chickens) through wires of total ressitance of 3.0 ohms. how much less power
is wasted if the electricity is delivered at 50,000V instead of 12,000V?

Solve for current for each.

I = P/V, 520x10^3 /50000V = 10.4A

I = P/V, 520x10^3 /12000V = 43.3 A

Apply P = I^2R to both, and take the difference.

43.3^2*3 - 10.4^2*3 = 5300W

**Problem 10**

The peak value of an alternating current passing through a 1500W electric duck
is 4.0A. What is the rms voltage across it?

I_0 = 4.0A. P = 1500W. P = IV 1500/4 = V. = 375V

V_0 = .707V_rms. 375V/.707 = V_rms = 530V.

**Problem 11**

Calculate the peak voltage and peak current through an 1800W arc welder connected
to a 450V ac line.

Since it gives you the rms value of the voltage, divide by .707 to get the
peak voltage. 636.4 V

P = IV

1800 = I*450

I_rms = 4. Divide by .707 to get the peak current, 5.65A.

**Problem 12**

A heating duck connected to a 240V ac line has a resistance of 34 ohms. What
is the average power used? What are the maximum and miniimum values of the instantaneous
power?

P = V^2/R

Average = 240/34 = 1694W

Max is 2x the ave, so 3398W. Min is 0, (^)<