CBSE Class 12-science Answered
Kindly solve it .
Asked by abhitailor158 | 08 Apr, 2020, 08:13: PM
Expert Answer
To get total current through 8Ω resistance by Superposition principle, we need to get individual contribution of currents
due to voltage source and current source independently.
First, as shown in figure-1, let us open the current source to get current IV due to voltage source.
It can be seen from figure 2, combination of resistors 24Ω , 4Ω and 20Ω is replaced by its equivalent resistance.
Equivalent resistance of combination of above mentioned resistors can be calculated as 11.67Ω.
Hence IV = 100 / ( 11.67+8) = 5.084 A
Next to get the current IC due to current source alone, as shown in fig.3, let us short the voltage source.
Current distribution in the circuit is considered as shown in fig.3
Voltage VBD across points B and D by considering the current passing through right side resistors is given by
VBD = 4 i2 + 8( i1 + i2 - i3 ) = 4 i2 + 8 ( 6 - i3 ) ...................(1)
in above eqn. we used the junction rule for current at point B.
Similarly, voltage VBD across points B and D by considering the current passing through left side resistor is given by
VBD = 24 i1 = 24( 6 - i2 ) = 144 - 24 i2 ...................(2)
By equating eqn.(1) and eqn.(2) , we get after simplification, 7i2 - 2i3 = 24 ............... (3)
Let us apply Kirchoff's voltage law to the loop A-C-B-A
- ( i1 - i3 ) 20 + 4 i2 - 24 i1 = 0
- 44 i1 + 4 i2 + 20 i3 = 0
-44 ( 6 - i2 ) + 4 i2 + 20 i3 = 0
Above equation is simplified as, 12 i2 + 5 i3 = 66 ........................(4)
By solving equations (3) and (4), we get i2 = 4.271 A , i3 = 2.95 A
current i1 = 6 - i2 = 6 - 4.271 = 1.729 A
Hence current IC through 8Ω resistor due to current source = i1 + i2 - i3 = 6 - 2.95 = 3.05 A
By superposition principle, current through 8Ω resistor = IV + IC = 5.084 + 3.05 = 8.134 A
Answered by Thiyagarajan K | 09 Apr, 2020, 09:51: AM
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