Question

For the circuit below, the switch has been open for a long time and is then closed at t=0. a. Make an accurate sketch of the voltage on R; versus time, labeling key values. (18 pts) b. Given that the time constant for this circuit is 80 ms, give the value of the voltage on R; at = 20 ms (7 pts) Switch R 1 k2 20 V - Rik2 c 100 uF R; 3 Κά The voltage and current waveforms shown below were measured across either a series RC or series RL load. First shown peak of v(t) at 10 us and of i(t) at 95 us, with exactly three periods shown. a. Is it RC or RL, and why? (7 pts) b. Find the value of R, and the value of either or L. (18 pts) 751 2.SE (1) -2.5H - 7.5 -- 10 1x10 2x10 3x10 Noteins, vt) in V. & ift) in ma An electron moves through a uniform magnetic field given by B = B_{ +38_j. At a particular instant, the electron has a velocity given by i = (40f +20) m's and the magnetic force acting on it is (8 x 10-18N)k. Find B. (25 pts)

Answer 1

1)

a) The circuit is kept open for quite some time and then closed at t=0. The circuit shown can be considered to be equivalent to R-C circuit, where the equivalent value of R can be calculated from the series and parallel connection.

When the switch is closed, the voltage in the circuit in R3 begins to grow and is shown in the figure below.

1 v (obopron) Vo you n Time 16)

b) Time constant of RC circuit=RC= 80*10^-3s(given)

Voltage at time t V(t) is related to voltage at time 0 V(0) by the relation:-

V(t)=V(0)(1-e^-t/RC)

Thus at t=20*10^-3s

V(t)=V(0)(1-e^-20/80)

where V(0)=20 V(given)

Thus, V(t)=20*(1-e^-0.25)

V(t)=4.42 V

2)

a) The voltage and current waveforms of a circuit are shown, with current peak at 95 $\mu$ s and Voltage peak at 10 $\mu$ s. Since current lags behind the voltage in the circuit, it is a RL circuit.

b) Here, time period T=1*10^-4s(for the waveforms)

Angular frequency $\omega$ =6.28/T=6.28*10⁴rad/s

Further, V/I=5/(0.01)=500(from the graph)

V=IR

Thus, resistance R=V/I=500 ohm

Also, V=IX_L

where X_L=impedance of RL circuit=$\omega$L

Thus, inductance L=V/I$\omega$=8*10^-3 H

3) charge on an electron q=1.6*10^-19C

magnetic field= B_xi+3B_xj

where i and j are unit vectors along X and Y directions respectively

velocity v=40 i+20 j m/s

Now, Force is given as F=q(v$\times$B)

=1.6*10^-19(40 i+20 j)$\times$(B_xi+3B_xj)=1.6*10^-19*100B_x k(along Z-direction)

=1.6*10^-17B_x N

The force is given as 8*10^-18N

Thus, equating, 1.6*10^-17B_x=8*10^-18

B_x​​​​​​​=0.5 Tesla