Maybe nobody ever mentioned it to you, but it turns out that
current is another one of those things that's always conserved ...
it can't created or destroyed, just like energy and mass.
The total current in a circuit is always the same, but it can get
split up and travel through different paths for a while.
<span>==> The total current is just the amount of current
that's flowing in and out of </span><span>the battery.
Diagram #1).
</span>The total current coming out of the battery is 15 A.
That current is going to split up when it reaches the resistors.
Part of it will flow through each resistor, but both of them
will still add up to 15 A .
You have 9 A flowing through one resistor.
So the current in the other resistor is (15 - 9) =<span> 6 A.
Diagram #2).
</span>The total current coming out of the battery is 10 A.
That current is going to split up when it reaches the resistors.
Part of it will flow through each resistor, but all of them
will still add up to 10 A .
You have 2.5 A through one resistor and 3.5 A through another one.
So the amount left for the last resistor is (10 - 2.5 - 3.5) =<span> 4 A.</span>
This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.
If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.
The energy dissipated by the circuit is the energy delivered by
the battery. We'd know what that is if we knew I₁ . Everything that
flows in this circuit has to go through R₁ , so let's find I₁ first.
-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
-- I₁ is 10volts/10Ω = 1 Ampere.
-- R1: 1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .
-- The battery is 10 volts.
7 of the 10 appear across R₁ .
So the other 3 volts appear across all the business at the bottom.
-- R₂: 3 volts across it = V₂.
Current through it is I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.
-- R3 + R4: 6Ω in the series combination
3 volts across it
Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere
-- Remember that the current is the same at every point in
a series circuit. I₃ and I₄ must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.
-- R₃: 1/2 Ampere through it = I₃ .
1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt
-- R₄: 1/2 Ampere through it = I₄
1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts
Notice that I₂ is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches I₁ coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.
Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up. But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.
The power supplied by the battery = (voltage) · (current)
= (10 volts) · (1 Amp) = 10 watts .
"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.
(30 minutes) · (60 sec/minute) = 1,800 seconds
(10 joules/second) · (1,800 seconds) = 18,000 joules in 30 min
The power (joules per second) dissipated by each individual resistor is
P = V² / R
or
P = I² · R ,
whichever one you prefer. They're both true.
If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total.
They're not asking for that. But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.
In an uniformly accelerated motion, the velocity of the object follows the law:
where
is the initial velocity, a the acceleration and t the time.
In our problem, the robot starts from rest, so the initial speed is zero:
. The robot is in free fall, so the acceleraion is the gravitational acceleration
. therefore, after a time
, the velocity is
To solve this problem we will apply the concepts related to the conservation of momentum. This can be defined as the product between the mass and the velocity of each object, and by conservation it will be understood that the amount of the initial momentum is equal to the amount of the final momentum. By the law of conservation of momentum,
Here,
= Mass of Basketball
= Mass of Tennis ball
= Initial velocity of Basketball
= Initial Velocity of Tennis ball
= Final velocity of Basketball
= Final velocity of the tennis ball
Replacing,
Solving for the final velocity of the tennis ball
Therefore the velocity of the tennis ball after collision is 11 m/s
Answer:
v = 3.6 m/s
Explanation:
given,
Inclination of the ramp, θ = 33⁰
Distance of the rope, d = 1.70 m
initial speed = ?
using conservation of energy
h = d sinθ
h = 1.7 x sin 33° = 0.926 m
moment of inertia of the solid ball
we know, v = r ω
v = 3.6 m/s
Hence, initial speed of the solid ball = 3.6 m/s
