Question

A current of 0.25 A flows in the main circuit. Now, the resistance R is disconnected and then connected across the $4\,\Omega$ resistance. The current in the circuit now is

• A. 6/19 A
• B. 2/9 A
• C. 4 A
• D. 1/2 A

Answer 1

Answer: (A)

Here 0.25 =$\frac{2}{\left(\frac{2R}{2 + R} \right) +4 +3}$ i.e. R = 2 $\Omega$
In the second case, I =$\frac{2}{2+ \left(\frac{2 \times 4}{4 +2} \right) + 3} = \frac{6}{19} A$