Well, first of all, you really shouldn't use ' W ' for the unit when you
talk about resistors.
You may have seen the resistors written as 6ω, 12ω, and 2ω in your
book or on the homework sheet. But that little symbol ' ω ' is not a ' w '.
It's the small Greek letter 'omega'. The CAPITAL omega is ' Ω '. It's used
to label resistors because it's short for "ohms". So the resistors in this
problem have resistances of 6Ω, 12Ω, and 2Ω, and we have to do some
manipulating of the individual resistors to find out what resistance the
battery actually sees.
The parallel combination of the first two resistors looks like a single
resistor, whose value is
1 / (1/6 + 1/12)
= 1 / (2/12 + 1/12)
= 1 / (3/12)
= 12/3 = 4Ω .
Now, that parallel combination is connected in series with 2Ω .
All three resistors together look like a single resistor of
4Ω + 2Ω = 6Ω .
So the battery thinks there's a single resistor connected to it,
with 6Ω of resistance. The current out of the battery is
I = V / R = (24v) / (6Ω) = 4 Amperes.
That 4 Amperes of current will split between the parallel resistors,
but it will ALL flow through the series 2Ω resistor because there's
no other path through that part of the circuit.
So the current through the 2Ω resistor is 4 Amperes. (B).
Note:
The POWER dissipated by the 2Ω resistor is
P = I² R = (4A)² · (2Ω) = 32 watts .
This is a fair amount of heat, so you'll need to provide some way
to remove the heat from the resistor, otherwise it'll burn or crack.
Answer:
(a) x=ASin(ωt+Ф₀)=±(√3)A/2
(b) x=±(√2)A/2
Explanation:
For part (a)
V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)
Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3
x=ASin(ωt+Ф₀)=±(√3)A/2
For part(b)
U=0.5E and U+K=E→K=0.5E
E=K(Max)
(1/2)mv²=(0.5)(1/2)m(Vmax)²
V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4
x=±(√2)A/2
