Question
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.
a) Calculate the current delivered by the battery.
(b) Calculate the potential difference across the load.
(c) Calculate the current in R2.
Answers
bobpursley
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/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
Henry2
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.
!/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.