5 pt (8c27p73) A controler on a electronic arcade game con- sists of a variable

Question

5 pt (8c27p73) A controler on a electronic arcade game con- sists of a variable resistor connected across the plates of a 0.310 F capacitor. The capacitor is charged to 5.40 V, then discharged through thhe resistor. The time for the potential difference across the plates to decrease to 0.800 V is mea- sured by a clock inside the game. If the range of discharge times that can be handled effectively is from 10.0pas to 4.00 ms, what should be the resistance range of the resistor? Low- end resistance? (in ohms) 10. AO 1.32x 101 BO 1.49 x 01 CO 1.69 x 101 DO 1.91 x 10 EO 216 x 101 FO 2.44 x 10 GO 2.75 × 10' HO 3.11 × 101 5 pt High-end resistance? (in ohms) 11. AO 5.78 × i03 BO 6.76 x i03 C0 791 x o? FO 1.27 × 104 DO 9.25 × 103 GO 1.48 × i04 EO 1.08 × 104 HO 1.73 × 104

Explanation / Answer

Given

Capacitance C = 0.310*10^-6 F

battery potential difference is V = 5.40 V initially

later due to the resistor the capacitor discharges through the resistor

so that the potential difference decreased to 0.80 V

the initial charge is Q0 = C*V = 0.310*10^-6*5.40 C = 1.674*10^-6 C

later the charge is q(t) = C*V(t) = 0.310*10^-6*0.800 C = 2.48*10^-7 C

for the discharging of a capacitor we have the expression in RC circuit is  

10.

q(t) = Q0(e^(-t/RC))

2.48*10^-7 = 1.674*10^-6*e^(-(10*10^-6)/(R*0.310*10^-6))

solving for R = 16.8931 ohm = 1.69*10^1 ohm

and

11.

at time t = 4 ms ,  

2.48*10^-7 = 1.674*10^-6*e^(-(4*10^-3)/(R*0.310*10^-6))

solving for R = 6757.234 ohm = 6.76*10^3 ohm

the resistance range  

low end resistance is R = 1.69*10^1 ohm ------> (C)

High end resistance is R = 6.76*10^3 ohm ------>(B)

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