According to the superposition principle, what percentage of the total current flowing through
the 5 Ohm resistance in the circuit below from the 5V source? What percentage of the power
supplied to the 5 Ohm is supplied by the 5V source?
Solution
Apply superposition:
Case1 : when 5V source is present. Then Short circuit the 15 V source
then Rnet = 10 + 5||10 = 13.33 ohms
net current = 5/13.33 = 0.375 A
current through 5 ohms = 10/(10+5) * 0.375 = 0.25 A
Power = I^2 R =0.3125 W
Case2 : when 15V source is present. Then Short circuit the 5 V source
then Rnet = 10 + 5||10 = 13.33 ohms
net current = 15/13.33 = 1.125A
current through 5 ohms = 10/(10+5) * 1.125 = 0.75 A
Power = 0.75^2 * 5 = 2.8125 W
thus total current through 5 ohms = 0.25 + 0.75 = 1 A
25 % of total current flowing through 5 ohms is supplied by 5 V source
Total Power = 2.8125 + 0.3125 = 3.125
thus 10% of total power supplied to 5 ohms is supplied by 5 V source
the 5 Ohm resistance in the circuit below from the 5V source? What percentage of the power
supplied to the 5 Ohm is supplied by the 5V source?
Solution
Apply superposition:
Case1 : when 5V source is present. Then Short circuit the 15 V source
then Rnet = 10 + 5||10 = 13.33 ohms
net current = 5/13.33 = 0.375 A
current through 5 ohms = 10/(10+5) * 0.375 = 0.25 A
Power = I^2 R =0.3125 W
Case2 : when 15V source is present. Then Short circuit the 5 V source
then Rnet = 10 + 5||10 = 13.33 ohms
net current = 15/13.33 = 1.125A
current through 5 ohms = 10/(10+5) * 1.125 = 0.75 A
Power = 0.75^2 * 5 = 2.8125 W
thus total current through 5 ohms = 0.25 + 0.75 = 1 A
25 % of total current flowing through 5 ohms is supplied by 5 V source
Total Power = 2.8125 + 0.3125 = 3.125
thus 10% of total power supplied to 5 ohms is supplied by 5 V source
