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  • The battery's internal resistance is negligible and the capacitance of the capacitor is  Then the st

The battery's internal resistance is negligible and the capacitance of the capacitor is 0.2 \mu \mathrm{F} Then the steady-state current in the
2 \Omega resistor is:  
  

Option: 1

0.9A


Option: 2

1.2A


Option: 3

0.7A


Option: 4

1.5A


Answers (1)

best_answer

The resistance of the parallel combination of 2\Omega and 3 \Omega resistors is given by
 \frac{1}{R}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6} \quad \Rightarrow R=1.2 \Omega
This resistance is in series with 2.8 \Omega giving a total effective resistance = 1.2 + 2.8 \Omega = 4 \Omega. In the steady state, charge on the capacitor C has stablised and hence no current passes through 4\Omega resistor which is in series with the capacitor. Thus the current through the circuit = 6 / 4=1.5 \mathrm{~A}
V_{A B}=1.5 \times 1.2=1.8 \mathrm{~V}$, I through $2 \Omega$ resistor $=1.8 / 2=0.9 \mathrm{~A}

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Irshad Anwar

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