# RC circuit equation why have negative sign

I'm reading a book named The Art of Electronics.

What are some of the features of circuits with capacitors? To answer this question, let’s begin with the simple RC circuit (Figure 1.31). Application of the capacitor rules gives

$$C{dV\over dt} = I = - {V\over R}$$

please see below picture:

According equation(1.19), I can solve its solution is

$$V = Ae^{-t/RC}$$

                                          or


$$V = {V_0}e^{-t/RC}$$

For equation(1.19) my understanding is that regardless of whether the current direction in the circuit is clockwise or counterclockwise, for both the capacitor and the resistor, the current flowing through them is in the same direction, so the signs are the same.

So in my opinion, equation(1.19) should be $$C{dV\over dt} = I = {V\over R}$$

I can't understand is why equation(1.19) add negative sign?

• When you connect a resistor and a capacitor in parallel as in the figure, does the voltage across the capacitor increase exponentially, getting endlessly higher at a rate that increases with the voltage? Or does it decrease exponentially towards zero at a rate proportional to the instantaneous voltage? What does that say about the sign of dV/dt with respect to the sign of V? Commented Jul 19 at 13:23
• @hobbs I'm new to electrical engineering. To be honest, I'm not sure how to answer your question specifically. Is there a possibility of an explanation from a definition (related to circuitry) or a mathematical perspective as to why there are signs in the equations?
– Tom
Commented Jul 19 at 13:46
• It's a question of which way you consider the current to be flowing. $I = \frac{V}{R}$ for current flowing into the resistor. $I = C\frac{dV}{dt}$ for current flowing into the capacitor. But current flowing into the resistor is flowing out of the capacitor, so when you're talking about the same current for both, the sign changes for one of them. Commented Jul 19 at 13:50
• @DaveTweed I think the current direction in this circuit can only be in one direction, either clockwise or counterclockwise.
– Tom
Commented Jul 19 at 14:44
• Exactly. And the direction you pick determines the correct sign for each component. Commented Jul 19 at 14:46

They should have drawn the current direction.

These equations match the drawing below. These are the definitions of a capacitor and resistor, current flows into the positive end of the voltage.

$$\I = C\frac{dV}{dt}\$$

$$\I = \frac{V}{R}\$$

Since the circuits are in series with both referenced to ground, one current must be negative. If the current is CW, then the capacitor current is negative. If the current is CCW, then the resistor current is negative.

simulate this circuit – Schematic created using CircuitLab

• In your circuit, I think current direction of capacitor should be toward up, according to your circuit, current flow out from positive of capacitor, then flow in positive of resistor.
– Tom
Commented Jul 19 at 14:33
• @Tom - the drawing represents the definitions of the components. If KVL is opposite this, then the current is negative. I clarified my answer. Commented Jul 19 at 16:57
• I'm sorry, I still don't understand the picture you drew, but analyzing equation(1.19) using KCL or KVL should be a good approach.
– Tom
Commented Jul 20 at 5:09

According to the conservation of energy $$P_{in}+P_{out}=0$$ According to passive sign convention power leaving a circuit is considered positive, while power entering a circuit is considered negative.

The capacitor is supplying energy from its stores to the circuit while the resistor is removing energy as heat so: $$P_{C}+P_{R}=0$$ $$v_Ci_C+v_Ri_R=0$$ $$-v_Ci_C=v_Ri_R\text{ OR }v_Ci_C=-v_Ri_R$$ Take your pick.

If you choose $$\i_C=i_R=i\$$ then either the resistor voltage or the capacitor voltage is negative.

If you choose $$\v_C=v_R=v\$$ then either the resistor current or the capacitor current is negative.

The passive sign convention would have the capacitor voltage or the capacitor current as negative. Generally if current leaves an element through its positive terminal, as all sources do, it is considered negative Hence $$\i_c\$$ is chosen as negative. The voltage could just as easily be chosen, but be consistent. So I choose: $$-i_c=i_R$$

$$-C\frac{dv_C}{dt}=\frac{v_R}{R}$$

Since they are in parallel $$\v_C=v_R=v\$$ and moving the sign to the other side then $$C\frac{dv}{dt}=\frac{-v}{R}$$

Actually the passive sign convention actually starts with power supplied into a circuit is negative, while power leaving a circuit is positive. Resistors always remove power so the VI product requires voltage and current to be both be positive or both be negative.

Capacitors can ether remove into storage or supply by releasing stored energy. As a source therefore the VI product must be negative. Either the voltage or the current can be negative, but not both.

So pick the polarities that you prefer consistent with the power convention.

• In your answer "The capacitor is acting as a source" is according with intention of author of the book, but i can't understand how you get −iC = iR, I think this is important, it can derive equation(1.19), can you explain more detail? Thank you.
– Tom
Commented Jul 19 at 14:19
• Conservation of energy. Pin+Pout= 0. I'll modify my answer. Commented Jul 19 at 14:28
• Thank you for you update your answer, now for me time is too late, I need to go sleep, tomorrow I will see your answer.
– Tom
Commented Jul 19 at 15:04