A) 0.6 A
B) 0.9 A
C) 1.2 A
D) 1.5 A
Correct Answer: B
Solution :
[b] Given Capacitance of the capacitor, C = 0.2 \[\mu F\] and e.m.f. of cell, E = 6V. Reactance of a capacitor for a cell, which is a DC source, is infinity. Therefore, no current a flows in \[4\Omega \] resistance. Resistances \[2\,\,\Omega \] and \[3\,\,\Omega \] (both in upper arm) are connected in parallel. Therefore, their equivalent resistance \[(R')=\frac{2\times 3}{2+3}=1.2\Omega \] Now, R' and \[2.8\,\,\Omega \] are in series combination. Therefore, equivalent resistance of the circuit, \[R=R'+2.8=1.2+2.8=4\,\Omega \] Current drawn in the circuit, \[I=\frac{E}{R}=\frac{6}{4}=1.5A\] Therefore, potential difference across \[2\,\,\Omega \] resistance, \[V=IR=1.5\times 1.2=1.8\,V\] Thus, current in \[2\,\,\Omega \] resistance \[({{I}_{1}})=\frac{V}{2}=\frac{1.8}{2}=0.9A\]
You need to login to perform this action.
You will be redirected in
3 sec
