Question:hard
The relation between the input current \((I)\) and the output voltage \((V)\) of a circuit is governed by the equation: \(C\dfrac{dV}{dt}=I(t)-m(t)\). The circuit is excited by \(I(t)=q\,\delta(t)\), where \(q\) is a real valued constant. \(V\) at \(t=0^-\) is \(V_0\).
Which of the following is an equivalent representation of the above case?
The relation between the input current \((I)\) and the output voltage \((V)\) of a circuit is governed by the equation: \(C\dfrac{dV}{dt}=I(t)-m(t)\). The circuit is excited by \(I(t)=q\,\delta(t)\), where \(q\) is a real valued constant. \(V\) at \(t=0^-\) is \(V_0\).
Which of the following is an equivalent representation of the above case?
Which of the following is an equivalent representation of the above case?
Show Hint
A current impulse of area \(q\) into a capacitor \(C\) causes an instantaneous voltage jump of \(q/C\); add this jump to the pre-impulse voltage \(V_0\) to get the new starting condition.
Updated On: Jul 20, 2026
- \(C\dfrac{dV}{dt}=-m(t)\), with \(V(t=0^-)=V_0+q/C\)
- \(C\dfrac{dV}{dt}=-m(t)\), with \(V(t=0^-)=V_0+q/C+m(t=0^-)\)
- \(C\dfrac{dV}{dt}=-m(t)\), with \(V(t=0^-)=V_0-q/C+m(t=0^-)\)
- \(C\dfrac{dV}{dt}=-m(t)\), with \(V(t=0^-)=q/C\)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Think in terms of charge instead of the differential equation.
A capacitor's charge and voltage are linked by $Q=CV$. The current $I(t)=q\,\delta(t)$ is not an ordinary current profile, it is a burst that dumps a fixed total charge $q$ onto the capacitor in essentially zero time, since the area under a delta function of strength $q$ is exactly $q$.
Step 2: Find the charge on the capacitor just before the burst.
Just before $t=0$, the capacitor voltage is $V_0$, so the stored charge is
\[ Q(0^-)=CV_0 \]
Step 3: Add the impulse charge.
The impulse adds charge $q$ almost instantly, and in that same vanishing instant the $m(t)$ term, which is not impulsive, cannot move any extra charge, since it needs a nonzero time interval to do so. So right after the burst,
\[ Q(0^+)=Q(0^-)+q=CV_0+q \]
Step 4: Convert the new charge back to a voltage.
\[ V(0^+)=\frac{Q(0^+)}{C}=\frac{CV_0+q}{C}=V_0+\frac{q}{C} \]
Step 5: State the equivalent system.
After the burst, the current source term is spent, so the governing equation returns to $C\,dV/dt=-m(t)$, now starting from the bumped-up voltage found above.
\[ \boxed{C\frac{dV}{dt}=-m(t),\quad V_{\text{start}}=V_0+\frac{q}{C}} \]
A capacitor's charge and voltage are linked by $Q=CV$. The current $I(t)=q\,\delta(t)$ is not an ordinary current profile, it is a burst that dumps a fixed total charge $q$ onto the capacitor in essentially zero time, since the area under a delta function of strength $q$ is exactly $q$.
Step 2: Find the charge on the capacitor just before the burst.
Just before $t=0$, the capacitor voltage is $V_0$, so the stored charge is
\[ Q(0^-)=CV_0 \]
Step 3: Add the impulse charge.
The impulse adds charge $q$ almost instantly, and in that same vanishing instant the $m(t)$ term, which is not impulsive, cannot move any extra charge, since it needs a nonzero time interval to do so. So right after the burst,
\[ Q(0^+)=Q(0^-)+q=CV_0+q \]
Step 4: Convert the new charge back to a voltage.
\[ V(0^+)=\frac{Q(0^+)}{C}=\frac{CV_0+q}{C}=V_0+\frac{q}{C} \]
Step 5: State the equivalent system.
After the burst, the current source term is spent, so the governing equation returns to $C\,dV/dt=-m(t)$, now starting from the bumped-up voltage found above.
\[ \boxed{C\frac{dV}{dt}=-m(t),\quad V_{\text{start}}=V_0+\frac{q}{C}} \]
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