The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors \[3\Omega ,9\Omega \]and \[9\Omega \] and a capacitor of 5.0 \[\mu F.\] |
How much is the current I in the circuit in steady state? [JEE ONLINE 12-04-2014] |
A) 1.6 A
B) 0.67 A
C) 2.5 A
D) 0.25 A
Correct Answer: B
Solution :
[b] |
In steady state capacitor is fully charged hence no current will flow through line 2. |
By simplifying the circuit |
Hence resultant potential difference across resistances will be 8.0 V. |
Thus current\[I=\frac{V}{R}\] |
\[=\frac{8.0}{3+9}=\frac{8}{12}\]or\[I=\frac{2}{3}=0.67A\] |
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