Question

In the circuit shown in Figure, calculate the following:
a. Potential difference between points $a$ and $b$ when switch $S$ is open.
b. Current through $S$ in the circuit when $S$ is closed.

Answer 1

i. Here $I_1=\frac{36-0}{9}=4A$

Also $I_2=\frac{36-0}{9}=4A$

Also, $36-V_a=I_1\times 6=6I_1$

$=6\times 4=24V$

$\therefore V_a=36-24=12V$

$36-Vb=3\times 4=12V$

and $V_b=36-12=24V$

ii. In loop $ABCIHA$

$-6(I-I_1)+3I_1=0$ or $-6I+6I_1+3I_1=0$

or $9I_1=6I$

$\therefore I=\frac{9}{6}{I}_1=\frac{3}{2}{I}_1$ (i)

In loop $CDEJIC$,

$-3(I-I_1+1_2)+6(I_1-I_2)=0$

or $-3I+3I_1-3I_2+6I_1-6I_2=0$

or $-3I+9I_1-9I_2=0$ (ii)

In loop $EFGHIJE$,

$36-3I_1-6(I_1-I_2)=0$

or $9I_1-6I_2=36$ (iii)

After solving Eqs. (i), (ii), and (iii), we get

$I_2=3Afrombtoa$.