A 2.2K Ohm resistor is placed in a series with 1.7K Ohm and 2.6K Ohm resistors i

Question

A 2.2K Ohm resistor is placed in a series with 1.7K Ohm and 2.6K Ohm resistors in a parallel. The entire circuit is driven by 18V. What is the voltage across each element? What is the current through each element?

* individual voltage dropped = source voltage(individual resistance/total resistance)
I used this equation to find the voltage across each element and calculated: 6.09V for 2.2KOhm, 4.70V for 1.7KOhm, and 7.20V for 2.6KOhm. Then I used ohm's equation V=I(R) to find the current through each element: all three came out to 2.7 6mA. I want to know if I have solved this problem correctly. If not, the display of steps would be greatly appreciated. Thank You!

Explanation / Answer

   Let R1 = 1.7 k ohm and R2 = 2.6 k ohm are in parallel.    The resultant of R1 & R2 is, R = R1R2/R1+R2 = 1.03 K     The voltage across circuit is. V = 18 v    The current through R1, i1 = V/R1 = 18/1.7 * 10^3 = 0.011 A    The current through R2, i2 = V/R2 = 18/2.6 * 10^3 = 0.007 A     The third R3 = 2.2 k ohm is in series with the both.    So the voltage at R3, is V3 = VR3 / (R+R3) = 12.26 V    Now the current at R3, i3 = V3/R3 = 0.0012 A    So the voltage across R1 = V1 = 18 v         The voltage across R2 = V2 = 18 V         The voltage across R3 = V3 = 12.26 V    The current at R1 = i1 = 0.011 A    The current at R2 = i2 = 0.007 A    The current at R3 = i3 = 0.0012 A    

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