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CBSE Class 12-science Answered

Please solve the following question using Kirchoff's current law and Kirchoff's voltage law asap. :))
Asked by mfirdous111 | 24 Nov, 2020, 12:05: AM
Expert Answer
Figure shows the current passing through each resistor. 
 
Since all resistors are in kΩ, we consider all currents in mA
 
By applying Kirchoff's voltage law (KVL) we get , 8 I1 + 4 I2 = 12 V  or  2 I1 + I2 = 3  .....................(1)
 
In loop-2, by applying KVL , we get , 10 I4 + 5 I5 = 4 I2 ..................(2)
 
In loop-3 , by applying KVL , we get , 18 I7 = 5 I5 ........................(3)
 
At node-Q , if we apply Kirchoff's Current law ( KCL ) , we get , I4 = I5 + I7 ...............(4)
 
Using eqn.(4) and eqn.(3), we rewrite eqn.(2) as 10 I5 + 28 I7 = 4 I2  ...................(5)
 
Using eqn.(3) , we rewrite eqn.(5) as ,  64 I7 = 4 I2   or   I7 = (1/16) I2 .................(6)
 
Using eqn.(3) , we get I5 = (18/5) I7 = (18/5) (1/16) I2 
 
Hence we get  I5 = (9/40) I2  ...........................(7)
 
From eqn.(4), we get , I4 = [ (9/40) + (1/16) ] I2 = (23/80) I2 .......................(8)
 
At node-P , if we apply Kirchoff's Current law ( KCL ) , we get , I1 = I2 + I4 ...............(9)
 
Let us rewrite eqn.(1) , using eqn.(9) as , 3I2 + 2I4 = 3   
 
  3 I2 + 2 (23/80) I2 = ( 143/40 ) I2 = 3     or   I2 = ( 120/143 )  = 0.839 mA   .....................(10)
 
Hence from eqn.(1) , we get I1 = (3/2) - (1/2) I2 = (3/2) - (1/2)(0.839) = 1.080 mA   ................(11)
 
Current I4 = (23/80) ( 0.839) = 0.241 mA
 
Current I5 = (9/40) ( 0.839) = 0.189 mA
 
Current I7 = (1/16) (0.839) = 0.052 mA
 
By knowing all the currents passing in the circuit , it is possible to get voltage across each resistor.
Answered by Thiyagarajan K | 24 Nov, 2020, 10:41: AM

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