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...
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...
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...
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...