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The circuit shown here has two batteries of 8.0 ~V and 16.0 ~V and three resistors 3 , 9 and 9 and a capacitor of 5.0 F . How much is the current I in the circuit in steady state?

Options

  1. A1.6 ~A
  2. B0.67 ~A
  3. C2.5 ~A
  4. D0.25 ~A

Correct answer

B. 0.67 ~A

Step-by-step solution

In steady state capacitor is fully charged hence no current will flow through line 2. By simplyfing the circuit Hence resultant potential difference across resistances will be 8.0 ~V . Thus current I = V R = 8.0 3+9 = 8 12 or, I = 2 3 =0.67 ~A

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