Recent content by rhemmin

  1. R

    Time required to charge capacitor to percentage of max capacitor voltage

    Voltage across R1 should equal the voltage source because of KVL. So, voltage across R1 should be 6V. Therefore, the current through the R1 equals 6/(8.9*10^-3) = .674mA. After removing capacitor and replacing voltage supply with short, resistance across the points where capacitor connects...
  2. R

    Time required to charge capacitor to percentage of max capacitor voltage

    Homework Statement The switch in the circuit is opened at t = 0. Determine the time required for the capacitor to charge to 53% of its maximum voltage in milliseconds. Given that R1 is 8.9 kΩ, R2 is 3.1 kΩ, C is 5.1 μF and Vs is 6 volts. Round your answer off to two decimals. Homework...
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    Equivalent resistance with possible short circuit

    Wow. Thank you both very much for the help. My first big problem was that I assumed there was a node where the wires crossed. For sure, I'll pay much closer attention to nodes from now on. Also, redrawing the diagram with the R=8.3k resistor going around the other circuit elements made the...
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    Equivalent resistance with possible short circuit

    Homework Statement Find the current flowing through the voltage source in mA. Assume Vs = 3.9 V and R = 8.3 kohm. Round off your answer to two decimal points. Homework Equations 1/R(equivalent) = 1/R1 + 1/R2 + ... I=V/R Kirchoff's voltage law The Attempt at a Solution I considered the 3...