A) Rate of energy loss in the source is = 8 W
B) Rate of energy conversion in the source is 16 W
C) Power output in is = 8 W
D) Potential drop across R is = 16 V
Correct Answer: A
Solution :
\[i=\frac{12}{(4+2)}=2A\] Energy loss inside the source \[={{i}^{2}}r\]\[={{(2)}^{2}}\times 2=8W\]You need to login to perform this action.
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