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For the attached circuit, i would be replacing voltmeter with 12V Light source and my intention is even after switch is off, light should stay for some time and dim later.

Will this circuit work for dimming?

Resistance of the bulb, either V^2/W (the square of the voltage divided by the power it consumes, in watt ) either about 15 times what you measure with a ohm-meter at room temperature (tungsten resistance increases with temperature).

Replace the actual voltmeter (used as an Amp-meter, here) with the resistance of the value you just get in first step.

Get started with the switch closed and open it after 5RC seconds (here, 5 x 1000 x 1000 ... well, replace C1 with one of 1mF, that will make 5 seconds), and then, open the switch. Observe how long it takes for Q1 to block the 12 V of the source, on the curve of voltage Vec of Q1 through time (Vec = voltage between the emitter and the collector). I would have used the NPN in the reverse direction, but since we are using it as a switch, it (seems to) work in the reverse active mode too as well (unless I missed a point).

5 kF cap, if it is really 5 kiloFarad, are known to not be linear, though, so the exact laps of time that the light bulb will stay on is only theoretical.

12V bulb is indifferent as circuit is closed with 12V supply as long as base current in transistor is active.

TO maintain the base current at transistor, after the switch is off, capacitor is expected to provide current till it drains. I have used 1000uF capacitor. Pls suggest what capacitor i shold ideally use and what formula should i apply to know the capitence of capacitor to be used.

The 12V bulb is indifferent but you need a resistance, not a "voltmeter", to run the simulator.

You can try a simulator with a given capacitance, run the simulation to get a time domain graph and see when the transistor cut-off.

Technically, you can try the formula for a linear capacitor to discharge. If it is charged at time t=0 to V_C0 then, through time, the cap C in serial with a resistor R in a closed loop (with nothing else in the loop) will discharge accordingly to V_c = V_C0 e^-t/RC.

With V_C0 = 12 and the final voltage around 0.7 volt, we end up with 0.7 = 12 e^-t/RC, or -ln(0.7 / 12) = 2.84 = t/RC. Given the time t that you want, in seconds, say 30, and R in Ohm, say 1000, I find C = 0.0105 F. Since capacitor are generally not really precise (tolerance), the real value may differ. I would try 10mF.