Question

Photocell can be used to charge capacitors as shown in the Figure 3, where a diode is placed
to prevent the capacitor from discharging. You were given FOUR new 10.00 mF capacitors. By
using all the capacitors given, how would you arrange the capacitors so that the total
capacitance of the combination can be fully charged by 25.00 hours? Sketch your answer in
the dotted area of the circuit.

Answer 1

R = 1.800 MΩ

The constant of capacitor = RC

Total charging time = 5RC

25 hr = 5RC

$25 \times 60 \times 60 = 5 \times 1.8 \times 10^6 \times C$

$C = \frac{25 \times 60 \times 60}{5 \times 1.8 \times 10^6}$

C = 10 mF

This is the capacitance of the combination.

$hυ = \frac{1}{2}mν^2 + hυ_0$

$\frac{1}{2}mν^2 = h(υ – υ_0) = v_0e$

λ = 150 nm

$υ_0 = 5.53 \times 10^{14} \; Hz$

$υ_0 = \frac{C}{λ}$

$υ_0 = \frac{3 \times 10^8}{150 \times 10^{-9}} = 20 \times 10^{14}\; Hz$

$v_0e = 6.63 \times 10^{-34}(20 - 5.53)\times 10^{14}$

$v_0 = \frac{9.58 \times 10^{-19}}{1.6 \times 10^{-19}}$

$v_0 = 6\;V$

Charging voltage of capacitor:

$V_c = V_s(1 - e^{-\frac{Z}{RC}})$

$V_c = 6(1 - e^{\frac{8 \times 60 \times 60}{3 \times 10^3 \times 10}})$

$V_c = 6(1 - e^{0.96})$

$V_c = 3.7 \;V$