A battery (ε=6.20 Vr =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.
To answer these questions, we'll need to use Ohm's Law and the rules for calculating current and resistance in parallel circuits.
(a) To calculate the current delivered by the battery, we can use Ohm's Law (V = IR), where V is the voltage and R is the total resistance of the circuit.
First, we need to calculate the total resistance of the parallel circuit. In a parallel circuit, the total resistance (R_total) can be calculated using the formula:
1/R_total = 1/R1 + 1/R2 + 1/R3
Plugging in the given values:
1/R_total = 1/6.00 + 1/9.00 + 1/18.0
Now we can solve for R_total:
1/R_total = 0.1667 + 0.1111 + 0.0556
1/R_total = 0.3333
R_total = 1/0.3333
R_total ≈ 3.00 Ω
Now, we can calculate the current using Ohm's Law by dividing the voltage (ε) by the total resistance (R_total):
I = ε/R_total
I = 6.20 V / 3.00 Ω
I ≈ 2.07 A
Therefore, the current delivered by the battery is approximately 2.07 Amperes (A).
(b) To calculate the potential difference across the load, we can use Ohm's Law again. The potential difference (V_load) across the load is equal to the current (I) multiplied by the resistance (R_total):
V_load = I * R_total
V_load = 2.07 A * 3.00 Ω
V_load ≈ 6.21 V
Therefore, the potential difference across the load is approximately 6.21 Volts (V).
(c) To calculate the current in resistor R2, we can use Ohm's Law again. The potential difference across resistor R2 is equal to the current (I2) multiplied by the resistance (R2):
V2 = I2 * R2
Rearranging the equation, we can solve for I2:
I2 = V2 / R2
The potential difference across the load (V_load) is the same as the potential difference across resistor R2 (V2). Therefore, we can substitute V_load for V2:
I2 = V_load / R2
Plugging in the values:
I2 = 6.21 V / 9.00 Ω
I2 ≈ 0.69 A
Therefore, the current in resistor R2 is approximately 0.69 Amperes (A).