Physics

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.

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  1. 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

  2. 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.

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