Answer:
a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes
b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
Explanation:
<u>Solution :</u>
(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation in the next form
P=∈*I-I^2*r (1)
Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by
I=∈/R+r (2)
When R is very small R << r, therefore the term R+ r will equal r and the current becomes
I= ∈/r
Now let us plug this expression of I into equation (1) to get the consumed power
P=∈*I-I^2*r
=I(∈-I*r)
=0
The consumed power when R is very small is zero
(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
I=∈/R
The dissipated power due toll could be calculated by using equation.
P=I^2*r (3)
Now let us plug the expression of I into equation (3) to get P
P=I^2*R=(∈/R)^2*R
=∈^2/R
Answer:
Explanation:
This is the row echelon form of A. This means that only two of the vectors in our set are linearly independent. In other words, the first two vectors alone will span the same subspace of as all three vectors.
Therefore, the linearly independent spanning set for the subspace is
Answer:
1.439 × 10¹¹ m is the answer.
Explanation:
1 light-second = 2.998 × 10⁸m
480 light-seconds = 480 × 2.998 × 10⁸
= 1439.04 × 10⁸
= 1.439 × 10¹¹
∴ 1.439 × 10¹¹ m is the answer.
The answer is 21m because the motion is in one dimension with constant acceleration.
The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>
Plugging in our known values, we have
<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>