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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. :))
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Asked by mfirdous111 | 24 Nov, 2020, 12:05: AM
answered-by-expert 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 I= (23/80) ( 0.839) = 0.241 mA
 
Current I= (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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