The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
Answer: Pieces of minerals, rocks, plant and animal remains.
Explanation: Pieces of minerals, rocks, plant and animal remains. whether or not patterns cause flow rates of rivers to vary. sand settles from faster-moving water;smaller costs of silt and clay that form up mud settle from slower moving water.
Potential energy is energy that is found in a system, grounded on the position of objects. The Coulomb (C) is the unit of charge, and the unit of electric potential is the Volt (V), which is equivalent to (J/C) or Joule per Coulomb.So the formula for this is potential = kQ / d, plugging in the given from the questions will give us:potential = 8.99e9N·m²/C² * 1.602e-19C / 0.053e-9m = 27 V