Answer
Final Answer
Element A delivers power
Element B delivers power
Element C absorbs power
Element D absorbs power
Element E absorbs power
Work Step by Step
Part a)
- Assuming the correct directions/polarities, power is being absorbed if the current enters (heads toward) the positive pole of the voltage
- Assuming the correct direction/polarities, power is being delivered if the current exits (heads outwards) the positive pole of the voltage
(note: by right direction/polarities we mean the the current/voltage has a positive sign; not the negative sign current)
- Or, another way to think of this is if the product of the voltage and current (the power) is positive, power is being delivered; otherwise, it is being absorbed
For element A:
assume $V_A$ with its positive pole heading up. Apply KVL
$-V_A-3+10+5=0$
$V_A=-3+10+5=12\ V$
Now assume a current $I_A$ that enters the positive pole of $V_A$.
First, find the $I_A$ current by applying KCL
$i_A+3+2=0$
$I_A=-3-2=-5\ A$
Since $I_A$ is negative sign, it means the direction is opposite. Therefore, correct assumption is letting $I_A$ exist the positive pole of $V_A$. Therefore, element A delivers power
If we find the power across element A
$P_A=-I_A\times V_A=-(-5)\times 12 = 60\ W$
for element B:
since we know that $I_A$'s correct direction is heading outwards from the positive pole of the $3\ V$, element B delivers power
if we calculate the power
$P_B=-5\times 3=15\ W$
for element C:
Again. $I_A$'s correct direction is heading toward the positive pole of the $5\ V$. Therefore, power is absorbed
$P_C=5\times 5=-25\ W$
for element D:
element D is in parallel with element E; therefore, they have same voltage (you can check by applying KVL}
$V_D=10\ V$
($V_D$ polarity matches that of element E)
current $3A$ enters the positive pole of voltage $V_D$' thus, the power is being absorbed
$P_D=-3\times 10=-30\ W$
element E:
current $2A$ heads towards the positive pole of the $10V$ voltage; thus, it absorbs power
$P_E=-2\times 10=-20\ W$
Part b)
Yes, the conservation of power is satisfied. We can prove this by the power
$$\Sigma P_{delivered} +\Sigma P_{absorbed}=0$$
$$\Sigma P_{delivered} =-\Sigma P_{absorbed}$$
$$P_A+P_B=-(P_C+P_D+P_E)$$
$$60+15=-(-25-30-20)$$
$$-75=-75$$
since left-hand side and right-hand side are equal, conservation is satisfied.
