A battery (ε= 6.20V, r = 0.100Ω) is connected to three light bulbs in parallels (R1= 6.00Ω, R2= 9.00Ω, R3= 18.0Ω).
a) Calculate the current delivered by the battery.
(b) Calculate the potential difference across the load.
(c) Calculate the current in R2.
load resistance:
1/R= 1/6+1/9+1/18=(3+2+1)/18=6/18
R= 3 ohm
current= 12/(3.1)= ...
voltage lost on internal resistance: Ir= 12*.1/3.1 = ....
voltage across load: 12-voltage in internal resistanc
Current in R2= voltage across load/R2
1/Rl = 1/R1 + 1/R2 + 1/R3.
!/Rl = 1/6 + 1/9 + 1/18 = 3/18 + 2/18 + 1/18 = 6/18 = 1/3.
Rl = 3 Ohms = Load Resistance.
a. I = E/(r+Rl) = 6.2/(0.1+3) = 2A.
b. Vl = I * Rl = 2 * 3 = 6 Volts.
c. I2 = Vl/R2.
To calculate the current delivered by the battery and the potential difference across the load, you can use Ohm's Law and the rules for combining resistors in parallel.
(a) Calculate the current delivered by the battery:
Using Ohm's Law (V = IR), we can determine the total resistance of the parallel circuit. The total resistance is given by the formula:
1/RTotal = 1/R1 + 1/R2 + 1/R3
Substituting the given values, we have:
1/RTotal = 1/6.00 + 1/9.00 + 1/18.0
Calculating the sum:
1/RTotal = 0.1667 + 0.1111 + 0.0556
1/RTotal = 0.3334
RTotal = 1/(0.3334)
RTotal ≈ 3.0 Ω
Now we can use Ohm's Law to calculate the current (I) delivered by the battery:
I = V/RTotal
I = 6.20V / 3.0Ω
I ≈ 2.07 A
Therefore, the current delivered by the battery is approximately 2.07 Amperes.
(b) Calculate the potential difference across the load:
Since the battery is connected in parallel with the lights, the potential difference across the load is the same as the potential difference across the battery, which is equal to the battery's emf (ε).
Therefore, the potential difference across the load is 6.20V.
(c) Calculate the current in R2:
Since the potential difference across the load and the battery is the same, the current passing through each resistor in parallel is also the same.
So the current in R2 is equal to the current delivered by the battery, which we already calculated to be approximately 2.07 A.