In the given figure, each diode has a forward bias resistance of $30 \Omega$ and infinite resistance in reverse bias. The current $\mathrm{I}_{1}$ will be:

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Current Electricity

In the given figure, each diode has a forward bias resistance of

30Ω

and infinite resistance in reverse bias. The current

I_{1}

will be:

Answer: (D)

A forward-biased diode contributes

30 Ω

, so each conducting branch is

130 + 30 = 160 Ω

, while a reverse-biased diode opens its branch. Two diode branches are forward biased, so they combine in parallel to

\frac{160}{2} = 80 Ω

. Adding the

20 Ω

of the source branch gives

100 Ω

across

200 V

, so

I_{1} = \frac{200}{100} = 2 A

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option