Answer
element A delivers power and has $P_A=400\ W$
element B absorbs power and has $P_B=-40\ W$
element C delivers power and has $P_C=400\ W$
element D absorbs power and has $P_D=-10\ W$
element E absorbs power and has $P_E=-450\ W$
Total delivered power $=500\ W$
Total absorbed power $=-500\ W$
Work Step by Step
-Power is absorbed if current enters (arrow faces) the positive pole of the voltage
-Power is delivered if current leaves (arrow opposes) the positive pole of the voltage
for element A)
current leaves the positive voltage pole. Therefore, power is delivered
$P_A=100\times 4=400\ W$
for element B)
current enters the positive voltage pole. Therefore, power is absorbed
$P_B=10\times -4=-40\ W$
for element C)
current leaves the positive voltage pole. Therefore, power is delivered
$P_C=1\times 100=100\ W$
for element D)
Since voltage across element D is negative it means that the correct polarity is the opposite (i.e. the negative pole is on the top and positive is on the bottom). Therefore, after fixing the polarity, the current enters the positive pole of the voltage; thus, power is absorbed
$P_A=1\times( -10)=-10\ W$
for element E)
current enters the positive voltage pole. Therefore, power is absorbed
$P_A=-5\times 90=-450\ W$
Calculate total delivered power
$\Sigma P_{delivered}= P_A+P_C=400+100=500\ W$
Calculate total absorbed power
$\Sigma P_{absorbed }= P_B+P_D+P_E=-40-10-450=-500\ W$
