Three different size lights are running off a low voltage system. The resistance of the three lights are R1 =3 ohms, R2 = 9 ohms, and R3 = 9 ohms. They are all connected in parallel with a 5.0 V battery. The internal resistance of the battery is negligible.
What is the current through R1?
I=R/V
1.67A? Right?
correct. The entire 5V drop happens for each bulb, since they are in parallel.
I = E/R = 5/3___Amps.
To find the current through resistor R1, you can use Ohm's Law. Ohm's Law states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by the resistance (R) of the resistor. In this case, the voltage across the resistor is given as 5.0 V, and the resistance of R1 is 3 ohms.
So, the formula to find the current through R1 is:
I = V / R
Plugging in the values:
I = 5.0 V / 3 Ω
Now we can calculate the current through R1.