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r''E/SS' ADVg Pr Ksea C = Lav/A? m C. R, : L0 JUL 1?; : 2.0 ML RI 1) [50] The circuit shown in
r''E/SS' ADVg Pr "Ksea C = Lav/A? " m C. R, : L0 JUL 1?; : 2.0 ML RI 1) [50] The circuit shown in the diagram has a switch with two possible positions: 5' > A means points 3 and A are connected, while 5' > B means points 5' and B are connected. When you start the switch was in positiOn S > A for a long time. a) at time t = 0 you set the switch to position 8 > B. Use a gure to sketch the voltage across resistor R1 as a function of time for t = 0.37". Label the axes in your gure. Do not use a graphing calculator, and accuracy where exp (0) m 3 is sufcient for your sketch. b) Report the time constant T in seconds. c) After the switch was quite some time in this position (S > B for many times T) what is the voltage drop across the capacitor C, and what is the voltage drop across resistor R2? d) Now you move 5' to position 5 > A. You can reset the clock to t = 0. Sketch the voltage drOp across the capacitor C as a function of time. Be careful about the time constant! What is this constant in seconds? e) if you measure the voltage drop across resistors R1 and R2 at any (but the same) time, how do they compare (i.e., what is the ratio AVl/AVQ)? 1
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