A) \[\frac{1}{5}\,\,amp\]
B) \[\frac{1}{10}\,\,amp\]
C) \[\frac{1}{45}\,\,amp\]
D) \[\frac{1}{15}\,\,amp\]
Correct Answer: B
Solution :
Let \[R\] be the total resistance of the circuit then \[\frac{1}{R}=\frac{1}{30}+\frac{1}{30+30}=\frac{1}{30}+\frac{1}{60}=\frac{2+1}{60}\] or \[\frac{1}{R}=\frac{3}{60}=\frac{1}{20}\] So, \[R=20\Omega \] Now current\[=i=\frac{E}{R}=\frac{2}{20}=\frac{1}{10}\,\,amp\]
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