please help, I only have one attempt remaining! 7. Given a maximum voltage of 3.

Question

please help, I only have one attempt remaining!

7. Given a maximum voltage of 3.9491 Volts and a minimum voltage of -0.0256 Volts, then what is half the maximum voltage value?
1/2 Maximum Voltage Value =   Volts.
(Answer with four significant figures or N/A if there is no answer.)

8. Let's say you measure t1/2 to be 0.0094 seconds and the resistance in the circuit is 90 Ohms. Then what is the capacitance in micro Farads?
Capacitance =   micro Farads.
(Answer with three significant figures or N/A if there is no answer.)

t1/2 means that the voltage (or charge) of the system will increase to half more of what is left in a time equal to t1/2 seconds.
Therefore if a system is already at half charge (t1/2 seconds after starting) then after t1/2 more seconds the system will be charged to 50% plus half of 50%. That is 25% more, or 75% of the entire charge.Let's say that four t1/2's have gone by. That means that the charge (or voltage) is at (50% + 1/2*50% + 1/2*1/2*50% + 1/2*1/2*1/2*50%) = 93.75% of maximum charge.
Yikes! Now look in your manual for a more simple mathematical derivation of this concept.

Explanation / Answer

7. Corrected Maximum Voltage Value =3.9491+ 0.0256  =3.9747 V

1/2 Maximum Voltage Value =3.9747/2= 1.98735 V

8. The voltage in an RC circuit that is charging or discharging varies as:
V(t) = Vo * exp(-t/T)
where Vo = V(t=0), and T = R*C
R is the circuit resistance, and C is the capacitance.
At t = t_1/2, V(t_1/2) = Vo/2, then this implies that:
Vo/2 = Vo * exp((-t_1/2)/(R*C))
ln(1/2) = -(t_1/2)/(R*C))
R*C = -(t_1/2)/(ln(1/2))
C = (t_1/2)/(R*ln(2))
C = (0.0094 sec)/(90 ohm * ln(2))
F = 1.5068*10^-4 F = 150.68. microfarads

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