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

Question

A current of 0.25 A flows in the main circuit. Now, the resistance R is disconnected and then connected across the 4 Ω resistance. The current in the circuit now is (A) 6/19 A (B) 2/9 A (C) 4 A (D) 1/2 A

Tardigrade Admin · Accepted Answer

Here 0.25 =(2/((2R)2 + R ) +4 +3) i.e. R = 2 Ω In the second case, I =(2/2+ ((2 × 4)4 +2 ) + 3) = (6/19) A

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

6/19 A

2/9 A

4 A

1/2 A

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

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

6/19 A

2/9 A

4 A

1/2 A

Here 0.25 =(2+R2R​)+4+32​ i.e. R = 2 Ω In the second case, I =2+(4+22×4​)+32​=196​A

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

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