Question

The current $I$ in the circuit shown is

• A. 1.33 A
• B. Zero
• C. 2.00 A
• D. 1.00 A

Answer 1

Answer: (A)

The circuit diagram can be redrawn as

For the loop $ABCDA$,
$+2-2I^{\prime }+2-2I=0$ ...(i)
For the loop $ABFEA$,
$2-2I+2-2\left(I-I^{\prime }\right)=0$
$4-2I-2I+2I^{\prime }=0$
$4-4I+2I^{\prime }=0$
$2=2I-I^{\prime }$ ...(ii)
From Eqs. (i) and (ii), we get
$2=2I-I^{\prime },2=I+I^{\prime }$
$4=3I,I=4/3=1.33A$