A) 1A
B) 3A
C) 2A
D) 4A
Correct Answer: A
Solution :
The effective resistance of the combination is \[\frac{1}{{{R}_{p}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1\] \[\Rightarrow \] \[{{R}_{p}}=1\,\Omega \] Total resistance of the circuit \[=R={{R}_{p}}+r=1+1=2\,\Omega \] So, battery current is \[i=\frac{V}{R}=\frac{2}{2}=1A\]You need to login to perform this action.
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