Question

1. In the first circuit you build, how does changing the voltage in a circuit affect the current in the circuit?

Answer 1

1) The voltage and the current in a circuit are related by Ohm's law:
$V=I R_{eq}$ 
where
V is the voltage of the circuit
I the current
$R_{Eq}$ is the equivalent resistance of the circuit.
This can be rewritten also as
$I= \frac{V}{R_{eq}}$ (1)

We can see from this equation that there is a direct proportionality between the voltage and the current: so, as we increase the voltage, the current increases as well; vice-versa, if we decrease the voltage, the current decreases.

2) When we add more light bulbs in series, the equivalent resistance of the circuit increases, because each light bulb has its own resistance and the equivalent resistance of a series of n resistors is
$R_{Eq} = R_1 + R_2 + ... +R_n$


3 )Looking at Ohm's law (1), we see that since the equivalent resistance has increased, if the voltage is not changed then the current in the circuit is not constant: instead, it decreases, so every time we add a new light bulb the brightness of each bulb decreases.

4) When we add more light bulbs in parallel, the equivalent resistance of the circuit decreases. In fact, each light bulb has its own resistance, and the equivalent resistance of n resistors in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$
We see from the formula that every time we add a new light bulb (a new resistor), the sum on the right increases. This means that the term $\frac{1}{R_{eq}}$ increases, and this means the equivalent resistance $R_{eq}$ of the circuit is decreasing.

5) By looking again at Ohm's law (1), we see that in this case, if the voltage is kept constant, then the overall current in the circuit is increasing, because $R_{eq}$ has decreased.