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For the network shown below, the switch S is closed at t = 0 with the capacitor uncharged. The value of di(t) / dt at t = 0+ will be—
-
- 100 A/sec
- – 100 A/sec
- 1000 A/sec
- – 1000 A/sec
Correct Option: B
When capacitor opened
at t = 0+
| IC(0–) = | = | R | 100 |
= 0.1 Amp
at t = o+
⇒ 100 = 1000 IC + 1 / C ∫ICdt
After differentiating, we get
| 0 = 1000 | + | lc | dt | C |
Capacitor in short circuited
| -1000 | = | = | dt | C | 1 × 10 -6 |
| = | - | dt | 103 |
= – 100 Ampere.
Hence alternative (B) is the correct choice.
