# RC circuit with current source simulate this circuit – Schematic created using CircuitLab

Suppose I have a series RC circuit with a current source.

The function of the current provided by the current source is:

$$I(t) = \begin{cases} I_0 & (0\text{s}\leq t \ (\text{mod}\ 2) < 1\text{s}) \\ 0 & (1\text{s}\leq t \ (\text{mod}\ 2) < 2\text{s}) \end{cases}$$

According to the current function, voltage across resistor is $V_R=I_0R$ and voltage across capacitor is $V_C=\frac{\int_{0}^{t} I(t) dt}{C}=\frac{I_0t}{C}$, when $0\leq t\leq 1$.

I have two confusions regarding this circuit:

1. According to Kirchoff's voltage law we need: $V_R = V_C$ i.e. voltage drop across resistor is equal to voltage drop across capacitor.

That is, we need $\frac{I_0t}{C}=I_0R\ \implies t/C=R$, which is obviously not true since $R$ is constant. So, where am I going wrong?

I think that to make Kirchoff's voltage law hold I need to consider voltage drop across current source too. But, I'm not sure. Anyhow, are my expressions for $V_C(t)$ and $V_R(t)$ correct?

2. In this kind of a circuit the voltage across capacitor $C$ will keep increasing with time, according to the equation $\frac{\int_{0}^{t}I(t)dt}{C}$. Is this practically possible?

• "Vr = Vc " why ? – Long Pham Apr 7 '18 at 8:17
• it's is only correct for parallel RC circuit – Long Pham Apr 7 '18 at 8:18

Exactly this. There is a voltage across the current source as well. This voltage times current equates to power consumption of the circuit. A correct Kirchoff voltage law goes: $$Vi = Vr + Vc$$ where Vi is the voltage across the current supply.
• @LongPham I meant voltage across capacitor will keep increasing with time as it is $\int_{0}^{t} I(t) dt/C$. – user133614 Apr 7 '18 at 8:26