# Photocell in the circuit

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A bulb B having a resistance of 100 ohm converts 31 % of its electrical energy into radiation having wavelength less than 5000 A and with an average wavelength of 4000 A in this range. 10% of this radiation is incident on a photocell D of efficiency 20% and is converted into an electric current. The photo electric threshold of the material is 5000 A.
The photocell (D) is connected in parallel with a resistance R (100 ohm) and a capacitance C (=100 $\mu$F) and finally in series with the bulb B & a source of emf E = 180 V. The key is now closed.

Now we have to find the charge on C and power dissipiated in (both in steady state)

I am not getting any start, now I am able to understand the given solution in my book .