A battery (ε= 10.0V, r = 0.50Ω) is connected to three light bulbs in parallels (R1= 15.0Ω, R2= 21.0Ω, R3= 24.0Ω). Calculate the current in R3.
See previous post: 12-13-18. 5PM.
1/R = 1/15 + 1/21 + 1/24 = 0.156
so R = 6.41 Ohms
total resistance including internal of battery = 6.91 Oms
total current = V/6.91 = 10/6.91 = 1.45 amps
loss of voltage due to internal resistance = 1.45*0.50 = 0.723 volts
so useful voltage = 10 - 0.723 = 9.28 volts
current we want = 9.28/24 = 0.387 amps
but now they are in parallel :)
To calculate the current in R3, we can use Ohm's law and the concept of parallel resistance.
1. Calculate the total resistance of the parallel circuit:
The formula for calculating the total resistance (Rp) of a parallel circuit is:
1/Rp = 1/R1 + 1/R2 + 1/R3
Let's plug in the values:
1/Rp = 1/15 + 1/21 + 1/24
Now, calculate the inverse on both sides of the equation:
Rp = 1 / (1/15 + 1/21 + 1/24)
Calculate the value of Rp.
2. Apply Ohm's law to calculate the total current (I):
Ohm's law states: V = I * R, where V is the voltage, I is the current, and R is the resistance.
In this case, the battery voltage (V) is given as 10.0V.
So, I = V / Rp
Calculate the value of I using the obtained value of Rp.
3. Calculate the current in R3:
Since all the light bulbs are connected in parallel, the current (I) flowing through each of them will be the same.
Therefore, the current in R3 will be the same as the total current (I) calculated in step 2.
Use the value of I calculated in step 2 to determine the current in R3.
If you provide the values calculated in the steps above, I can help you with the final calculation.