1.

Initial voltage on capacitor VO as marked |VO| = 5 V, VS = 8 u (t), where u (t) is the unit step. The voltage marked V at t=0^+ is _____________(a) 1 V(b) -1 V(c) \(\frac{13}{3}\) V(d) –\(\frac{13}{3}\) VThis question was posed to me during an online exam.My question comes from Dot Convention in Magnetically Coupled Circuits in chapter Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

Correct option is (C) \(\frac{13}{3}\) V

Easiest explanation: Applying voltage divider method, we get,

I = \(\frac{V}{R_{EQ}} \)

= \(\frac{8}{1+1||1} \)

= \(\frac{8}{1+\frac{1}{2}} = \frac{16}{3}\) A

I1 = \(\frac{16}{3} × \frac{1}{2} = \frac{8}{3}\) A

And \(I’_2 = \frac{V}{R_{EQ}}= \frac{5}{1+1||1}\)

= \(\frac{5}{1+\frac{1}{2}} = \frac{10}{3}\) A

Now, \(I’_1 = I’_2 × \frac{1}{1+1} \)

= \(\frac{10}{3} × \frac{1}{2} = \frac{5}{3}\) A

Hence, the net current in 1Ω RESISTANCE = I1 + \(I’_1\)

= \(\frac{8}{3} + \frac{5}{3} = \frac{13}{3}\) A

∴ Voltage DROP across 1Ω = \(\frac{13}{3} × 1 = \frac{13}{3}\) V.


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