The circuit contains two diodes each with a forward resistance of 50 $$\Omega$$ and with infinite reverse resistance. If the battery voltage is 6 V, the current through the 120 $$\Omega$$ resistance is ______ mA

In the given circuit (see the figure in the question) the 6 V battery is connected in such a way that both diodes are forward-biased, while their reverse directions are never called upon; hence the reverse resistance (infinite) is irrelevant for the present polarity.

Whenever a silicon diode conducts, it is replaced by the forward-equivalent model
   • a constant junction drop of about $$0.7\;\text{V}$$, and
   • a dynamic (ohmic) forward resistance $$r_f = 50\;\Omega$$ (given in the statement).

Tracing the only closed path from the positive terminal of the battery back to its negative terminal, we encounter in succession:

$$\text{Battery }(6\;\text{V}) \;\rightarrow\; \text{Diode 1 }(0.7\;\text{V} + 50\;\Omega) \;\rightarrow\; 120\;\Omega \text{ resistor} \;\rightarrow\; \text{Diode 2 }(0.7\;\text{V} + 50\;\Omega) \;\rightarrow\; \text{back to battery.}$$

Thus the net series combination facing the battery is

Voltage drops due to junctions   : $$0.7 + 0.7 = 1.4\;\text{V}$$
Ohmic resistances in series  : $$50 + 120 + 50 = 220\;\Omega$$.

Applying Kirchhoff’s Voltage Law (sum of rises = sum of drops):

$$6\;-\;1.4 \;=\; I \times 220$$
$$\Rightarrow\; I \;=\; \frac{4.6}{220}\;\text{A}$$
$$\Rightarrow\; I \;\approx\; 0.0209\;\text{A} \;=\; 20\;\text{mA (to two-significant-figure accuracy).}$$

The same current flows through every element in this single-loop circuit, hence the current through the $$120\;\Omega$$ resistor is about $$20\;\text{mA}$$, matching the required answer.