A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V battery.
A) Calculate the voltage drop across each resistor in the circuit.
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`