a circuit contains a 9v battery and two 1 ohm bulbs. what is the voltage drop across each.

how are the bulbs connected?

It depends upon whether the bulbs are in series or in parallel.

If they are in series, each gets half the battery voltage; however that is not the way light bulbs are usually arranged.

Your teacher should know that.

To calculate the voltage drop across each bulb in the circuit, we need to apply Ohm's law.

Ohm's Law states that the voltage drop (V) across a resistor is equal to the current (I) flowing through it multiplied by the resistance (R):

V = I * R

First, we need to calculate the current flowing through the circuit. Since the bulbs are connected in series, the same current will flow through both bulbs.

To find the current (I), we can use Ohm's Law again, this time using the total resistance (R_total) of the circuit and the battery voltage (V_battery):

I = V_battery / R_total

In this case, the resistance of each bulb is 1 ohm, and since they are in series, the total resistance (R_total) will be the sum of the resistances of the bulbs, i.e., 1 ohm + 1 ohm = 2 ohms.

Now we can calculate the current:

I = 9V / 2Ω
I = 4.5 Amps

Since the same current flows through both bulbs, the voltage drop across each bulb will be the product of the current and its resistance:

V_bulb1 = I * R_bulb1 = 4.5 A * 1 Ω = 4.5 V
V_bulb2 = I * R_bulb2 = 4.5 A * 1 Ω = 4.5 V

Therefore, the voltage drop across each bulb in the circuit is 4.5 volts.