Answer Happy

2. Use your graph in 1(a) and the method described in figure 2 to find an average value for the time constant of the circuit. (5 marks) Time constant is the time required to reach 63.2% of initial value. capacitor will discharge 36.8% from graph 1 A. we have given equation of discharging where, 1/RC-0.17 t=5.88 seconds Average value of time constant=5.88 seconds 3. Use your answer in 2 and the capacitance of the capacitor to find a value for the resistance of the resistor. (3 marks) We know that time constant t = RC, T=0.5s C-4700 μF-4700x10 F 5.00 Re 1251.560 4700x10 4. Use the gradient of your graph in 1(b) and the capacitance of the capacitor to find a second value for the resistance of the resistor. (5 marks) H

1. (a) Plot a graph of V (y-axis) against time, f (x-axis), for your data. (b) Plot a graph of InV (y-axis) against time, t (x-axis), for your data. Chart Title 7 f(x) = 4,16260201356905 exp(-0.16953523735531 x) 5 10 15 20 25 30 voltage (V) voltage (V) 6 S 4 m 21 0 4 16 in 32 m 2 1 0 ***** 0 t(s) Chart Title f(x) = 4.16260201356905 exp(- 0,16953523735531 x) 10 15 20 25 30 5 t(s) 35 Se 40 99 45 45