The relation between the input current (I) and the output voltage (V) of a circuit is governed by the equation: 𝐶𝑑𝑉 𝑑𝑡 =𝐼(𝑡)−𝑚(𝑡)...

 

Q.  The relation between the input current (I) and the output voltage (V) of a circuit is governed by the equation: 𝐶𝑑𝑉 𝑑𝑡 =𝐼(𝑡)−𝑚(𝑡). The circuit is excited by I(t) = q(t), where q is a real valued constant. V at 𝑡=0− is V0. Which of the following is an equivalent representation of the above case?

Solution

The relationship between input current I(t) and output voltage V(t) is governed by the equation:

C dV/dt = I(t) − m(t)

The circuit is excited by an impulse current:

I(t) = qδ(t)

Given: V(0⁻) = V₀

Step 1: Substitute the input current

Since I(t) = qδ(t), substitute it into the circuit equation:

C dV/dt = qδ(t) − m(t)

Step 2: Integrate across the impulse

Integrate both sides over the interval from 0⁻ to 0⁺:

C ∫ dV/dt dt = q ∫ δ(t) dt − ∫ m(t) dt

Assuming m(t) is finite and has no impulse at t = 0, its integral over this infinitesimal interval vanishes. Also, the integral of the Dirac delta impulse is 1.

C [V(0⁺) − V(0⁻)] = q

Step 3: Calculate the voltage jump

Using the initial condition V(0⁻) = V₀:

C [V(0⁺) − V₀] = q

V(0⁺) − V₀ = q/C

V(0⁺) = V₀ + q/C

Step 4: Find the equation after the impulse

For t > 0, the impulse term δ(t) is zero. Therefore:

C dV/dt = −m(t)

Correct option

Option (A)

C dV/dt = −m(t)

V(0⁺) = V₀ + q/C

The impulse current produces an instantaneous voltage jump of q/C across the capacitor.

The capacitor voltage jumps by $q\mathrm{/}C$, while the ordinary signal $m(t)$ does not contribute a finite amount of charge over that infinitesimal interval.