A Kirchhoff Voltage and Current Law problem.

We can use voltage superimposition to solve this problem.

1. The voltage across $R_3$ due to $V_1$ , $V_{31} = \dfrac {R_2||R_3}{R_1+R_2 || R_3} V_1 = \dfrac{\frac{20\times 40}{20+40}}{10+\frac{20\times 40}{20+40}} \times 10 = \dfrac{40}{7}$
2. The voltage across $R_3$ due to $V_2$ , $V_{32} = \dfrac {R_1||R_3}{R_2+R_1 || R_3} V_1 = \dfrac{\frac{10\times 40}{10+40}}{20+\frac{10\times 40}{10+40}} \times 20 = \dfrac{40}{7}$

Therefore, the voltage across $R_3$ , $V_3 = V_{31}+V_{32} = \dfrac {80}{7} = \boxed{11.44}$ V.

Using Mesh Analysis, let $V$ be the voltage across the 40 ohm resistor.

$\frac{10-V}{10} = I_{1}$

$\frac{20-V}{20} = I_{2}$

Adding the two equations,

$I_{1} + I_{2} = \frac{1}{20} (40 - 3V)$

Since,

$V = 40 (I_{1} + I_{2} )$

Thus,

$V = 80 - 6V$

$V = \frac{80}{7} \approx \boxed{11.44}$

(10 - v)/ 10 + (20 - v)/ 20 = v/ 40

(40 - 4 v)/ 40 + (40 - 2 v)/ 40 = v/ 40

80 - 6 v = v

7 v = 80

v = 80/ 7 = 11.428571428571428571428571428571+

Checked that 0.428571 = 0.142857 + 0.285714 that the direction of current becomes opposite for R1.

The whole circuit presents an only set of voltages and currents.

Answer: 11.428571428571428571428571428571+

using kirchoff's law in the first loop containing v1 equation is 5i1+4i2=1 applying kirchoff's law in the second loop containing v2 equation is 2i1+3i2=1 solving both equation we have i1=(-1/7) i2=(3/7) current passing through 40 ohm resistor is 2/7 using ohm's law v=ir v=2/7x40=80/7