what is the voltage across the capacitor 0.1 seconds later? what is the current 0.1 seconds later?
The correct answer and explanation is:
To answer what is the voltage across the capacitor 0.1 seconds later and what is the current 0.1 seconds later, I need some more context, such as:
- The initial voltage across the capacitor
- The circuit setup (e.g., is it charging or discharging? Is it an RC circuit? What are the resistor and capacitor values?)
- The input voltage or source voltage, if any
- The time constant (or values of R and C to calculate it)
Assuming a simple RC discharging circuit:
Let’s say you have a capacitor CC initially charged to voltage V0V_0, connected through a resistor RR, and it starts discharging at time t=0t=0.
Formula for voltage across the capacitor at time tt:
V(t)=V0e−tRCV(t) = V_0 e^{-\frac{t}{RC}}
Formula for the current through the resistor at time tt:
I(t)=V0Re−tRCI(t) = \frac{V_0}{R} e^{-\frac{t}{RC}}
Explanation:
The voltage across a capacitor in an RC circuit changes exponentially with time during charging or discharging. The time constant τ=RC\tau = RC determines how fast this change happens.
- When discharging, the capacitor voltage decreases exponentially from V0V_0 toward zero.
- The current through the resistor also decays exponentially and is proportional to the voltage divided by resistance RR.
Example:
If
- V0=10 VV_0 = 10 \, V,
- R=1 kΩ=1000 ΩR = 1 \, k\Omega = 1000 \, \Omega,
- C=100 μF=100×10−6 FC = 100 \, \mu F = 100 \times 10^{-6} \, F,
then τ=RC=1000×100×10−6=0.1 seconds\tau = RC = 1000 \times 100 \times 10^{-6} = 0.1 \, \text{seconds}
At t=0.1 st = 0.1 \, s: V(0.1)=10×e−0.10.1=10×e−1≈10×0.3679=3.679 VV(0.1) = 10 \times e^{-\frac{0.1}{0.1}} = 10 \times e^{-1} \approx 10 \times 0.3679 = 3.679 \, V I(0.1)=101000×e−1=0.01×0.3679=3.679×10−3 A=3.679 mAI(0.1) = \frac{10}{1000} \times e^{-1} = 0.01 \times 0.3679 = 3.679 \times 10^{-3} \, A = 3.679 \, mA
Summary:
- Voltage across the capacitor 0.1 seconds later is approximately 3.68 V.
- Current through the circuit 0.1 seconds later is approximately 3.68 mA.
300-Word Explanation:
When a capacitor discharges through a resistor, the voltage across the capacitor decreases exponentially over time. The mathematical relationship describing this behavior is V(t)=V0e−t/RCV(t) = V_0 e^{-t/RC}, where V0V_0 is the initial voltage, RR is the resistance, CC is the capacitance, and tt is the time elapsed since discharge began.
The product RCRC is known as the time constant τ\tau. It represents the time it takes for the voltage (or current) to decrease to about 37% (1/e) of its initial value. For example, if τ=0.1\tau = 0.1 seconds, after 0.1 seconds the voltage will have dropped to roughly 37% of the starting voltage.
The current through the circuit during discharge is directly related to the voltage across the capacitor by Ohm’s Law I(t)=V(t)/RI(t) = V(t)/R. Since voltage decays exponentially, current does as well, following I(t)=V0Re−t/RCI(t) = \frac{V_0}{R} e^{-t/RC}.
Using typical values for resistance and capacitance allows calculation of exact voltages and currents at any time. For instance, if a capacitor initially charged to 10 volts discharges through a 1 kΩ resistor with a 100 µF capacitance, the time constant is 0.1 seconds. At t=0.1t = 0.1 seconds, the voltage drops to about 3.68 V, and the current falls to about 3.68 mA.
This exponential decay behavior is fundamental in timing circuits, filters, and many electronic applications where gradual voltage or current change is necessary. The rate at which voltage and current decrease is solely determined by the values of RR and CC.