A car battery with ? =12.0V and neglible internal resistance (r=0) is connected

Question

A car battery with ?=12.0V and neglible internal resistance (r=0) is connected three identical resistors with R= 20.0 ?.

Part A Find the current and power of each resistor consumes when the switch is open.

Part B Find the current and power of each resistor consumes when the switch is closed

Part C if the battery is old and the internal resistor of the battery is r=1.00 ohm, with the switch S closed, find the voltage applied on the resistor on the left and the power dissipated on the battery due to the internal resistance.

Explanation / Answer

A) When the switch is open, we have two resistors in series. The current through them is 12/ 40 = 0.3 A. Power of each is I^2*R, or 1.8 W. B) When the switch is closed, we have two resistors in parallel in series with a third resistor. The current through the third resistor is 12 / (20 + 20//20) = 12 / 30 = 0.4 A. The voltage through each parallel one is 10* 0.4 = 4 V, so the current through each is 4/20 = 0.2 A. The power through these can be computed by V^2/R, or I^2*R, as above. C) the internal resistance becomes another series resistor. The voltage applied on the left one becomes 20* (12/31) = 7.74 V. The power dissipated through the small resistance becomes (12/31)^2 * 1 = 0.15 W. :)

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