t conclusion on the time constant can be drawn from comparison of Measurement 2
ID: 1873785 • Letter: T
Question
t conclusion on the time constant can be drawn from comparison of Measurement 2 with Measurement 5? 2. Wha The circuit contains an ideal battery, two resistors and a capacitor (C-250F The switch is closed at time t-0, and the voltage across the capacitor is recorded as a function of time as shown in the graph. 3. Read the data in the graph carefully and answer the following questions Ci (a) What is the voltage of the battery? y9vol (b) what is the time constant for this circuit when the switch is closed? tas- V=V'('-L V- 90e 2-0 , V (c) what is the time constant for this circuit when the switch is open? tarn- (d) What is the resistance of the resistor R1? R (e) What is the resistance of the resistor R? R (0 What is the voltage across resistor Ri at t-2.0 seconds? Vi- (g) What is the voltage across resistor R2 at 1-2.0 seconds? y, (h) what is the total current produced by the battery at t-0? 2.4 mA 1) What is the current through resistor R, at t-5.0 seconds ? l 6) What is the charge on the capacitor t- 15.0 seconds? (k) When is the switch opened? 52 var 9 Volt 42Explanation / Answer
(i) From Graph, we see that potential across capacitor at t = 5 s is 7.5 V
As volatge of battery is = 9 Volts
Hence, voltage across resistance R1 must be = 9 - 7.5 = 1.5 V
R1 = 10,000 ohm
Hence, current through R1 = 1.5/10000 = 0.15 mA
(j) From Graph, we can see that, voltage across capacitor at t = 15 s is = 9 V
As Capacitance C = 250 x 10-6 F
Charge Q = CV = 9 x 250 x 10-6 C = 2.25 mC
(k) From Graph, we can see that switch is opened at t = 15 s
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