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The voltage across the capacitor, as shown in the figure, is expressed as
Vc (t) = A1 sin(ω1 t – θ1) + A2 sin(ω2 t – θ2)
The values of A1 and A2 respectively, are
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- 2.0 and 1.98
- 2.0 and 4.20
- 2.5 and 3.50
- 5.0 and 6.40
Correct Option: A
Opening current source,
| VC1(t) = | = | ||||||
| R + | jω1C | jω1RC + 1 | |||||
| jω1C | |||||||
| = | [ ω1 = 10 rad / sec ] | |
| 1 + j10 |
| = | , where θ1 = tan-1(10) | |
| √101 |
VC1(t) ≅ 2 sin (10t – θ1) ...(A)
Now, shorting voltage source, we have
| VC2(t) = | |||||
| R + | jω2C | ||||
| jω2C | |||||
| = | = | [ ω2 = 5 rad / sec ] | ||
| 1 + jω2CR | 1 + 5j |
| = | sin( 5t - θ2) , where θ2 = tan-1(5) | |
| √26 |
VC2(t) ≅ 1.97 sin(5t – θ2) ...(B)
From equation (A) and equation (B), we have
A1 = 2.0, A2 = 1.98
