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CBSE Class 12-science Answered

find current supplied by each battery and power dissipated in 1ohm resistor
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Asked by snehashiragannavar773 | 21 Oct, 2023, 02:38: PM
answered-by-expert Expert Answer
Let us apply Kirchoff's voltage law to the loop ABEFA . We get
 
-0.25 I2  +12 -12 +0.20 I1 = 0
 
From above circuit , we get ,  I1 = 1.25 I2 .................. (1)
 
Let us apply Kirchoff's voltage law to the loop BCDEB
 
begin mathsize 14px style I subscript 3 space minus 12 space plus space 0.25 space I subscript 2 space end subscript equals space 0 end style ......................... (2)
 
If we apply Kirchoff's curremt law to the juntion B , wegt ,
 
I3 = I1 + I2 .................... (3)
 
By using eqn.(1) and eqn.(3) , we rewrite eqn.(2) as
 
begin mathsize 14px style 2.5 space I subscript 2 space equals space 12 space space end style
 
Hence I2 = 4.8 A ;
 
From eqn.(1) , we get , I1 = 6 A ;
 
Hence current passing through 1 Ω resistor is I3 = ( 6 + 4.8 ) A \ = 10.8 A
 
Power diispation in 1 Ω resitor = I32 R = (10.8 × 10.8 × 1 ) W = 117 W
 
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