Q8. Two de voltage sources with EMF of 150V and 155V are connected in parallel. The series internal resistances of the voltage sources are 32 and 2.52 respectively. The above setup connected in parallel to a common load of 120 resistance. Using superposition theorem, determine: i Current supplied by cach voltage source. i. Power consumed by the load.

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Q8.
Two de voltage sources with EMF of 150V and 155V are connected in parallel. The
series internal resistances of the voltage sources are 32 and 2.52 respectively. The above
setup connected in parallel to a common load of 120 resistance. Using superposition
theorem, determine:
i. Current supplied by each voltage source.
ii. Power consumed by the load.
Transcribed Image Text:Q8. Two de voltage sources with EMF of 150V and 155V are connected in parallel. The series internal resistances of the voltage sources are 32 and 2.52 respectively. The above setup connected in parallel to a common load of 120 resistance. Using superposition theorem, determine: i. Current supplied by each voltage source. ii. Power consumed by the load.
Expert Solution
Step 1

Given,

 Two DC voltage source with EMF of 150 V & 155 V are connected in parallel.

Series internal resistance of the voltage sources are 3 Ω and 2.5 Ω respectively.

The above setup connected in parallel to a common load of 12 Ω resistance.

Step 2

(i) To find current supplied by each voltage source.

Circuit representation from given conditions,

Electrical Engineering homework question answer, step 2, image 1

As in superposition theorem the response in any element is the sum of the response obtained with one source acting at a time and other source being deactivated.

Deactivation means all the independent source are replaced by their internal resistance as voltage source by short circuit and current source by open circuit.

 

As from 150 V voltage source acting and other sources keep deactivating.

Electrical Engineering homework question answer, step 2, image 2

 

I1=150 V12Ω||2.5Ω+3Ω=72.5 A

Hence current supplied by 150 V voltage source is 72.5 A

 

Current through load of 12 Ω by current division,

 I112Ω=72.5×2.512+2.5=12.5 A

 

As from 155 V voltage source acting and other sources keep deactivating.

Electrical Engineering homework question answer, step 2, image 3

I2=155 V12Ω||3Ω+2.5Ω=31.632 A

Hence current supplied by 155 V voltage source is 31.632 A

 

Current through load of 12 Ω by current division,

I212Ω=31.632×312+3=6.3264 A

 

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