Question

how does a parallel circuit change when a branch is added?
the total resistance increases, so the current in the circuit decreases.
the total resistance decreases, so the current in the circuit increases.
the total resistance increases, so bulbs shine at a lower brightness.
the total resistance decreases, so bulbs shine at a higher brightness.

Explanation:

In a parallel circuit, the formula for total resistance \( \frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\dots+\frac{1}{R_n} \). When a branch (a resistor) is added, the number of terms in the sum increases, so \( \frac{1}{R_{total}} \) increases, meaning \( R_{total} \) decreases. By Ohm's law \( I = \frac{V}{R} \) (where \( V \) is the voltage, constant in parallel), a lower total resistance leads to a higher total current. For bulb brightness, in parallel, each bulb has the same voltage as the source. Adding a branch doesn't change the voltage across each bulb, so their brightness (related to power \( P=\frac{V^2}{R} \) for each bulb) remains the same. So we analyze each option:

• First option: Total resistance should decrease, not increase. Wrong.
• Second option: Adding a branch decreases total resistance (\( R_{total} \) smaller), so \( I=\frac{V}{R_{total}} \) (V constant) increases. Correct.
• Third option: Total resistance decreases, not increases. And bulb brightness doesn't change. Wrong.
• Fourth option: Bulb brightness depends on voltage across them (same as source), which doesn't change. So brightness doesn't increase. Wrong.

Answer:

The total resistance decreases, so the current in the circuit increases.