In the circuit shown, the current in the \[1\Omega \] resistor is: [JEE MAIN 2015] |
A) 0.13 A, from Q to P
B) 0.13 A, from P to Q
C) 1.3 A, from P to Q
D) 0A55
Correct Answer: A
Solution :
[a] |
\[{{E}_{eq}}=\frac{\frac{5}{5}\frac{6}{3}=\frac{9}{5}\times \frac{3}{3}}{\frac{1}{3}+\frac{1}{5}}=\frac{3}{\cancel{15}\times \frac{8}{\cancel{15}}}=\frac{3}{8}\] |
\[{{r}_{eq}}=\frac{3\times 5}{8}=\frac{15}{8}\] |
\[\] |
\[=0.13\] |
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