In the circuit shown in figure all the three diodes D1, D2, D3 have forward resistance $50\ \Omega$ each and infinite backward resistance. If the battery voltage is 5V, the current through $100\ \Omega$ resistance is

The positive terminal of the $5 \text{ V}$ battery is connected to the left side of the circuit, making diodes D1 and D2 forward-biased and diode D3 reverse-biased.

Since D3 has infinite backward resistance, it acts as an open circuit and no current flows through its branch.

The forward resistance of D1 and D2 is $50 \Omega$ each. The resistance of the top branch is $R_1 = 50 + 150 = 200 \Omega$.

The resistance of the middle branch is $R_2 = 50 + 50 = 100 \Omega$.

These two branches are in parallel, so their equivalent resistance $R_p$ is:

$R_p = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{200 \times 100}{200 + 100} = \dfrac{20000}{300} = \dfrac{200}{3} \Omega$

The $100 \Omega$ resistor is in series with this parallel combination. The total equivalent resistance of the circuit is:

$R_{eq} = R_p + 100 = \dfrac{200}{3} + 100 = \dfrac{500}{3} \Omega$

The total current $I$ from the battery, which is the current flowing through the $100 \Omega$ resistor, is:

$I = \dfrac{V}{R_{eq}} = \dfrac{5}{\dfrac{500}{3}} = \dfrac{15}{500} \text{ A}$

$I = 0.03 \text{ A} = 30 \text{ mA}$