In a parallel connection of resistors shown at the right, each has one lead conn

Question

In a parallel connection of resistors shown at the right, each has one lead connected to one common point (a) and the other lead connected to a second common point (b). Since all are R2 connected to the same two points, resistors in parallel have the same potential difference (voltage). Because the current is not "used up," the current 1 to the group of resistors splits up: 1 11 12 13. We can replace parallel resistors with one resistor with an equivalent resistance Rpar. This one resistor will pass the same total current and dissipate the same total power for a given potential difference. That is Vab Rpar VabVR1 Vab/R2 Vab/R3 Dividing by the common potential difference vab and generalizing to any number of resistors in parallel 1 1 1 1 (8) resistors in parallel) The battery shown has an emf of 240 V and an internal resistance of 2 Q. Calculate the power r dissipated in the 24 resistor using the following steps a) Calculate the R of the three parallel L resistors. 2/24 18 X240 240 l812 12amp V- 2 V Then redraw the circuit, replacing the three parallel resistors with one resistor of value Rpar Page 17 Physics 236, 252

Explanation / Answer

!/ R parallel = 1/24 + 1/144 + 1/144 = 1/18

Rparallel = 18 ohms

b) R series= 18 + 2 = 20 ohms

c) I = V/R = 240/ 20 = 12 A

Voltage drop across parallel resistors= 12 ( 18) = 216 V

power dissapiated = v^2/ R = 216^2 / 24=1944 watts

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