One of the ways to find the current through the resistor `R_2` is by considering the Kirchoff's Law and the junction rule for currents in the given circuit.
In order to write the Kirchoff's Law, let's assume that the current `I_1` through `R_1` is directed to the right, the current `I_2` through `R_2` is directed down, and the current `I_3` through `R_3` is also directed to the right.
Note that while the circuit is not drawn as closed, all the grounded branches of the circuit have the same potential (zero), so they could be considered connected in one point.
Then, for the left loop
`V_(s1) - I_1R_1 - I_2R_2 = 0` (1)
and for the right loop
`V_(s2) +I_3R_3 - I_2R_2 = 0` (2)
The junction rule for the currents: `I_1 = I_2 + I_3` (The current entering the junction equals the sum of the currents leaving the junction).
Plugging in the given values gives us the system of equations that can be solved for `I_2` :
`100I_1 +100I_2 = 10` (1)
`100I_2 - 100I_3 = 5` (2)
`I_1 - I_2 - I_3 = 0` (3)
Simplify the equation (1) by dividing both sides by 10:
`10I_1 + 10I_2 = 1`
Now, let's multiply the equation (3) by -10 and combine it with (1) in order to eliminate `I_1` :
`-10I_1 + 10I_2 + 10I_3 = 0`
Adding (1) and (3) results in
`20I_2 + 10I_3 = 1` (4)
Now solve this together with (2): `20I_2 - 20I_3 = 1` :
Multiply (4) by 2:
`40I_2 +20I_3 = 2` (4)
`20I_2 - 20I_3 = 1` (2)
Adding these equations together results in
`60I_2 = 3` , so this means `I_2 = 3/60 = 1/20 = 0.05 A` . Since the result is positive, the original choice of direction (down) was correct.
The current through resistor R2 is down and it equals 0.05 A.
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