Thermistor circuit (2)

Label node voltages $V_1$ and $V_2$ as shown in the modified diagram. For each of these nodes, the sum of the currents flowing out must equal zero.

$\frac{V_1 - V_S}{R_5} + \frac{V_1 - V_2}{R_1 + R_2} + I_S = 0 \\ \frac{V_2}{R_6} + \frac{V_2 - V_1}{R_1 + R_2} - I_S = 0$

Solve for $V_1$ and $V_2$ . The current through $R_1$ and $R_2$ is:

$I = \frac{V_1 - V_2}{R_1 + R_2}$

The voltage at $A$ is:

$V_A = V_1 - R_1 I$

The voltage at $B$ is:

$V_B = V_2 + R_4 I_S$

Numerical results:

$V_1 = 8 \\ V_2 = 4 \\ V_A = 7 \\ V_B = 8 \\ V_{AB} = -1$

By mesh analysis :

$\begin{aligned} I_1 & = 2 & ...(1) \\ -20 + 3I_3 + (0.5+1.5)(I_3-I_1) + 1I_3 & = 0 \\ -20 + 3I_3 + 2(I_3-2) + I_3 & = 0 \\ 6I_3 & = 24 \\ \implies I_3 & = 4 \ \rm mA \end{aligned}$

Therefore, $V_{AB} = 1.5(I_3-I_1) - 2I_1 = 1.5(4-2) - 2(2) = \boxed{- 1} \ \rm V$ .