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A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω...
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When the three resistors are connected in series, the equivalent resistance of the circuit is
So the equivalent resistance of the given circuit is ``
Then the current in the circuit, and through each of the resistors, is determined by
`` , where U is the voltage supplied by the battery.
(When the resistors are in series, the current through each resistor is the same.)
`i=(9V)/(35Omega) = 9/35 A`
The voltage drop across each resistor is the current through the resistor times resistance:
`U_1 = iR_1 =(9/35)*5=9/7 V`
`U_2=iR_2 = (9/35)*10=18/7 V`
`U_3=iR_3 = (9/35)*20=36/7 V`
Posted by ishpiro on May 27, 2013 at 7:53 PM (Answer #1)
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