Question:
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 \(V(t=0^-)=q/C\)
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The Correct Option is A
Solution and Explanation
Step 1: Write down what the impulse does physically.
The original equation is
\[ C\frac{dV}{dt}=I(t)-m(t)=q\,\delta(t)-m(t) \]
A current impulse \(q\,\delta(t)\) delivers a finite, instantaneous burst of charge \(q\) to the capacitor exactly at \(t=0\). Everywhere except at that single instant, the equation just reads \(C\,dV/dt=-m(t)\), since \(\delta(t)=0\) for \(t\neq0\).
Step 2: Integrate the equation across the impulse.
Integrate both sides from \(t=0^-\) to \(t=0^+\), an interval that shrinks to zero width but still contains the impulse:
\[ C\int_{0^-}^{0^+}\frac{dV}{dt}\,dt=\int_{0^-}^{0^+}q\,\delta(t)\,dt-\int_{0^-}^{0^+}m(t)\,dt \]
Step 3: Evaluate each term.
The left side is \(C[V(0^+)-V(0^-)]\) by the fundamental theorem of calculus. On the right, the sifting property of the delta function gives \(\int_{0^-}^{0^+}q\,\delta(t)\,dt=q\). Since \(m(t)\) is an ordinary, non-impulsive function, its integral over a vanishing time interval is zero:
\[ \int_{0^-}^{0^+}m(t)\,dt=0 \]
Step 4: Combine these results.
\[ C[V(0^+)-V(0^-)]=q-0=q \] \[ V(0^+)=V(0^-)+\frac{q}{C}=V_0+\frac{q}{C} \]
Step 5: Write the reduced equation.
For every instant after the impulse has acted, the current source term is gone, since \(\delta(t)=0\) for \(t>0\), leaving
\[ C\frac{dV}{dt}=-m(t) \]
This reduced equation needs its own starting value to be solved going forward, and that starting value is exactly the jumped voltage found in Step 4, that is \(V_0+q/C\).
Step 6: Rule out the other options.
Options (B) and (C) both add an extra \(m(t=0^-)\) term to the jump, but Step 3 showed the ordinary function \(m(t)\) contributes nothing to the jump, since it has no impulsive content there; only the delta-function current source produces a jump. Option (D) drops \(V_0\) entirely, which would only be correct if the capacitor started at zero voltage before the impulse arrived, which is not stated here.
Final Answer:
\[ \boxed{C\frac{dV}{dt}=-m(t),\ \text{starting from } V_0+\frac{q}{C}} \]
The original equation is
\[ C\frac{dV}{dt}=I(t)-m(t)=q\,\delta(t)-m(t) \]
A current impulse \(q\,\delta(t)\) delivers a finite, instantaneous burst of charge \(q\) to the capacitor exactly at \(t=0\). Everywhere except at that single instant, the equation just reads \(C\,dV/dt=-m(t)\), since \(\delta(t)=0\) for \(t\neq0\).
Step 2: Integrate the equation across the impulse.
Integrate both sides from \(t=0^-\) to \(t=0^+\), an interval that shrinks to zero width but still contains the impulse:
\[ C\int_{0^-}^{0^+}\frac{dV}{dt}\,dt=\int_{0^-}^{0^+}q\,\delta(t)\,dt-\int_{0^-}^{0^+}m(t)\,dt \]
Step 3: Evaluate each term.
The left side is \(C[V(0^+)-V(0^-)]\) by the fundamental theorem of calculus. On the right, the sifting property of the delta function gives \(\int_{0^-}^{0^+}q\,\delta(t)\,dt=q\). Since \(m(t)\) is an ordinary, non-impulsive function, its integral over a vanishing time interval is zero:
\[ \int_{0^-}^{0^+}m(t)\,dt=0 \]
Step 4: Combine these results.
\[ C[V(0^+)-V(0^-)]=q-0=q \] \[ V(0^+)=V(0^-)+\frac{q}{C}=V_0+\frac{q}{C} \]
Step 5: Write the reduced equation.
For every instant after the impulse has acted, the current source term is gone, since \(\delta(t)=0\) for \(t>0\), leaving
\[ C\frac{dV}{dt}=-m(t) \]
This reduced equation needs its own starting value to be solved going forward, and that starting value is exactly the jumped voltage found in Step 4, that is \(V_0+q/C\).
Step 6: Rule out the other options.
Options (B) and (C) both add an extra \(m(t=0^-)\) term to the jump, but Step 3 showed the ordinary function \(m(t)\) contributes nothing to the jump, since it has no impulsive content there; only the delta-function current source produces a jump. Option (D) drops \(V_0\) entirely, which would only be correct if the capacitor started at zero voltage before the impulse arrived, which is not stated here.
Final Answer:
\[ \boxed{C\frac{dV}{dt}=-m(t),\ \text{starting from } V_0+\frac{q}{C}} \]
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