Question

Charging characteristics for a series capacitive circuit is:

Vc=〖V(1-e〗^(-(t/T))), where T=CR
time is constant
Capacitor C= 100nF
Reisistor R=47kΩ
Supply Voltage, V= 5 Volts

1. Determine the value of (t) when vc =4.15 volts.
2. Differentiate the charging equation and find the rate of change of voltage at 6 ms

Answer 1

Answer on Question #78800 – Math – Differential Equations

Charging characteristics for a series capacitive circuit is

$$V _ {C} = V \left(1 - e ^ {- \frac {t}{T}}\right),$$

where $T = CR$, time is constant;

Capacitor, $C = 100 \, nF$;

Resistor, $R = 47 \, k\Omega$;

Supply voltage, $V = 5 \, \text{Volts}$.

Question

1. Determine the value of $t$ when $V_{C} = 4.15 \, \text{Volts}$.

Solution

$$V _ {C} = V \left(1 - e ^ {- \frac {t}{T}}\right)$$

Solve for $t$

$$\begin{array}{l} 1 - e ^ {- \frac {t}{T}} = \frac {V _ {C}}{V} \\ e ^ {- \frac {t}{T}} = 1 - \frac {V _ {C}}{V} \\ - \frac {t}{T} = \ln \left(1 - \frac {V _ {C}}{V}\right) \\ t = - T \ln \left(1 - \frac {V _ {C}}{V}\right) \\ t = - R C \ln \left(1 - \frac {V _ {C}}{V}\right) \\ \end{array}$$

Substitute

$$\begin{array}{l} t = - (47 \times 10^{3} \, \Omega) (100 \times 10^{-9} \, F) \ln \left(1 - \frac {4.15 \, \text{Volts}}{5 \, \text{Volts}}\right) \\ t = 0.00832820 \, s \approx 8.328 \times 10^{-3} \, s = 8.328 \, ms \\ \end{array}$$

Answer: $t = 8.328 \, ms$

Question

2. Differentiate the charging equation and find the rate of change of voltage at 6 ms.

Solution

$$V _ {C} = V \left(1 - e ^ {- \frac {t}{T}}\right)$$

Differentiate both sides with respect to $t$

$$\frac {d}{d t} (V _ {C}) = \frac {d}{d t} \left(V \left(1 - e ^ {- \frac {t}{T}}\right)\right)$$ $$\text{rate of change of voltage} = \frac{dV_C}{dt} = V \left(\frac{1}{T}\right) e^{-\frac{t}{T}} = \frac{V}{RC} e^{-\frac{t}{RC}}$$ $$\text{Capacitor, } C = 100 \, \text{nF}$$ $$\text{Resistor, } R = 47 \, \text{k}\Omega$$ $$\text{Supply voltage, } V = 5 \, \text{Volts}$$ $$t = 6 \, \text{ms}$$ $$\text{rate of change of voltage} = \frac{dV_C}{dt} = \frac{5 \, \text{Volts}}{(47 \times 10^3 \, \Omega)(100 \times 10^{-9} \, \text{F})} e^{-\frac{6 \times 10^{-3} \, \text{s}}{(47 \times 10^3 \, \Omega)(100 \times 10^{-9} \, \text{F})}} \approx 296.793 \, \text{Volts/s}$$

Answer: rate of change of voltage $= \frac{dV_C}{dt} = V \left(\frac{1}{T}\right) e^{-\frac{t}{T}} = \frac{V}{RC} e^{-\frac{t}{RC}}$

$$\text{rate of change of voltage} \mid_{t=6 \, \text{ms}} = \frac{dV_C}{dt} \mid_{t=6 \, \text{ms}} = 296.793 \, \text{Volts/s}$$

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