# 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 .

asked Jan 24, 2017
Please be more specific about what difficulty you are having with the solution given. Which line do you not understand?
Sorry, but as said I have not getting any start that is difficulty from first step.
If that is the case, I think you need to do some more studying. You can return to the problem later. If you cannot even make an attempt to solve the problem, you are not ready for it.
In this what they have done when we got i =9.1