in the fig the current supplied by the battery
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1/R=1/20+1/10
1/R=3/20
R=20/3
V=IR
2=I×20/3
2×3=I×20
6=I×20
6/20=I
I=0.3A
Current is 0.3 ampere
1/R=3/20
R=20/3
V=IR
2=I×20/3
2×3=I×20
6=I×20
6/20=I
I=0.3A
Current is 0.3 ampere
vaibhav9495:
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sir .1 is the correct answer and lower side reversed biased where current is not flow bt thanks i found it
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