Question

What is eqivalent charge when we add same resistance in series.

Answer 1

We will solve this problem by this way:

Let`s find eqivalent charge:

q₁=$I\times t=\frac{U}{R1}\times t$ , q₁-charge, I-current, t- spent time to flow current, R₁=$n\times R$ -total resistance, U-voltage.

If we add a same resistor to circuit, then:

(total resistance) R₂=$n\times R$ +$R$, this means that total resistance increases as we add new resistors and also it means :

q₂=$I\times t=\frac{U}{n*R+R}\times t$. In this case, charge decreases as resistance increases

q₂$\lt$q₁ because R₁$\lt$ R₂