Question

Section_ PHYS. I 2001 250 Lab Manual 2018 Name 31A- Experiment: AC Circuits D Circuit A: A sinusoidal voltage of constant amplitude V IV and varying frequency is applied to two resistors in series. For the circuit to the right,F sketch your anectlou Grnerater prediction for the amplitude of the voltage across the Ry resistor R/2). SHOW WORK! CIRCUITI as a function of frequency (Let R 2) Circuit B: A sinusoidal voltage of constant amplitude Vo- IV and varying frequency is applied across a resistor and capacitor in series. For the circuit to the right, sketch your prediction for the amplitude of the voltage across the capacitor as a function of frequency. The time constant is RC SHOW WORK! Generator 103

Answer 1

Here, R_2 = R_1/2 the applied voltage is sinusoidal i.e. = V_0 sin(wt) w = 2 pi f = angular Frequency F = linear Frequency Now, the amplitude of the sinusoidal voltage V_0 = 1V therefore Now, the voltage across the R_2 is By the Voltage division rule V_R_2(t) = V(t) times R_2/R_1 + R_2 V_R_2(t) = V_0 sin(wt) times R_2/R_1 + R_2 Now, R_2 = R_1/2 and V_) = 1V therefore V_R_2 (t) = R_2/R_! + R_2 sin(wt) \r\n V_R_2(t) = R_1/2/R_1 + R_1/2 Sin(wt) V_R_2(t) = R_1/3R_1 Sin(wt) V_R_2(t) = 1/3 sin(wt) therefore the amplitude of the voltage across R_2 will be A = 1/3 therefore the amplitude of the voltage is independent of frequency \r\n V_C(t) = voltage across the capacitor therefore By voltage division rule V_C(t) = V(t) times -j/wc/R - j/wc V_C(t) = V(t) times 1/jwc/R + 1/jwc thereforeV_C(t) = V(t) times 1/jwc/jRwc + 1/jwc therefore V_C(t) = V(t) times 1/1 + jRwc times (1 - jRwc)/(1 - jwRc)