Question

SERIES CIRCUIT:

In the circuit, the total voltage source is 30.0 V, R1 = 4.0 x 10^1 Ω and R2 = 15 Ω. After 10 seconds, how many coulombs of charge has passed through the circuit?

Answer 1

The figure shows a series circuit with two resistances and a voltage source. Since the circuit is series through the resistance flows the same current $I$. According to Ohm's law, the voltage drop across the resistances is $U_1=I\cdot R_1$ ; $U_2=I\cdot R_2$ According to Kirchhoff's voltage law, the sum of all voltages over any closed loop is . In our case, this means $U-U_1-U_2=0$ . From the equation one can determine the magnitude of the current $U=I\cdot (R_1+R_2)$;

$I=\frac{U}{R_1+R_2}=\frac{30.0V}{(40+15)\Omega}=0.55 A$.

By definition, the current is equal to the charge transferred per unit time, i.e. $I=\frac{\Delta Q}{\Delta t}$. Thus the charge equals $\Delta Q=I\cdot \Delta t=0.55 A\cdot 10s=5.5 C$ .

Answer: After 10 seconds, $5.5 C$ has passed through the circuit.