8. Analyze the circuit below. The value of Vm (on the graph)...

Transcript

00:01 So here in this portion we are given this diagram where this source of the battery is connected to a resistance in parallel.

00:09 Then this resistor is further connected with the inductor and the inductor is connected with the capacitor.

00:18 So here this is a current source it is represented by the i s of t that is equals to 0 .01 of course 10 raised to the power 4 of t.

00:29 This is in ampere here let's say this is a positive terminal here it is a negative terminal.

00:35 The value of the resistance is 1 kilo ohm where the arrow is showing the direction of the ir.

00:43 Inductance is given that is equals to 20 hertz where the arrow is showing the value of the current due to inductance and we are given the value of the capacitance that is equals to 0 .5 micro farad arrow showing the value of the ic at which it is moving.

00:59 So from here we are having the value of i s t that is equals to 0 .01 multiplied by the cos 10 raised to the power 4 of t.

01:12 It is an ampere we are given the value of w that is equals to 10 raised to the power 4 radian per second.

01:18 So as we know that xl is equals to omega l that is equals to 10 raised to the power 4 multiplied by the value of l that is equals to 20 multiplied by 10 raised to the power minus 3.

01:29 So solving from here we get the value of xl that is equals to 200 ohm.

01:34 Now in the same way we have to find out the value of xc that is equal to 1 divided by omega c is equals to 1 divided by 10 raised to the power 4 that is multiplied by 0 .5 multiplied by 10 raised to the power minus 6.

01:49 So solving from here the value of xc is equals to 200 ohm.

01:54 So here in this diagram we can write here z that this value from here is equals to 0 .01 to the angle 0 degree.

02:04 This here is equals to z of 200 and this one here is minus j of 200.

02:11 Now from here we can apply the nodal analysis at the node v from here.

02:24 So that from here is minus 0 .01 plus v divided by the 1 of k plus v divided by the j 200 plus v divided by the minus j 200 that is equals to 0...

Question

8. Analyze the circuit below. The value of $V_m$ (on the graph) is: 0.5 sin ?t $R_1$ + $R_2$ -1 V $V_{out}$ $V_m$ $V_{out}$ A. 1.75 V B. 1.10 V C. 1.625 V D. 0.875 V

Question

8. Analyze the circuit below. The value of $V_m$ (on the graph) is: 0.5 sin ?t $R_1$ + $R_2$ -1 V $V_{out}$ $V_m$ $V_{out}$ A. 1.75 V B. 1.10 V C. 1.625 V D. 0.875 V

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