Question

Figure shows a circuit with three ideal batteries in it. The circuit elements have the following values $ΔV_{B_1}=3.0V,ΔV_{B_2}=6.0V{R}_1=2.0Î\copyright ,R_2=4.0Î\copyright$ The currents $i_1,i_2$ and $i_3$ as shown in the circuit have the values

• A. 0.50 A, - 0.25 A, + 0.25 A
• B. 0.25 A, - 0.50 A, - 0.25 A
• C. 0.50 A, 0.50 A, 1,0 A
• D. - 0.25 A, - 0.50 A, 0.25 A

Answer 1

Answer: (A)

Given circuit
In $ABCD$, From Kirchhoffs law
$2i_1+2i_1−4i_2+3−6=0$
$4i_1−4i_2=3…$ (i)
and in $CDEF$ From Kirchhoffs law
$2i_3+4i_2+6−6+2i_3=0$
$4i_2+4i_3=0i_2=−i_3$ .. (ii)
$i_1+i_2=i_3…$ (iii)
From Eq. (ii), we get
$i_1+i_2=−i_2$
From Eq. (i), $i_1=−2i_2$
$3=4\left(−2i_2\right)−4i_i$
$3=−12i_2$
$i_2=−0.25A$
$i_1=−2i_2=0.50A$
and $i_3=−i_2=0.25A$