A circuit consists of a 9.0 v battery connected to three resistors (37 , 17 , and 110 ) in series. (a) find the current that flows through the battery. ma (b) find the potential difference across each resistor. v37 = v v17 = v v110 = v
(a) The equivalent resistance of three resistors in series is the sum of the three resistances: [tex]R_{eq}=R_1 + R_2 + R_3 = 37 \Omega + 17 \Omega + 110 \Omega = 164 \Omega[/tex] And so the current flowing through the circuit (and the battery) can be found by using Ohm's law: [tex]I= \frac{V}{R_{eq}}= \frac{9.0 V}{164 \Omega}=0.055 A [/tex]
(b) The potential difference across each resistor is given by Ohm's law: [tex]V_{37} = I R_{37}=(0.055 A)(37 \Omega)=2.04 V[/tex] [tex]V_{17}=I R_{17}=(0.055 A)(17 \Omega)=0.94 V[/tex] [tex]V_{110}=I R_{110}=(0.055 A)(110 \Omega)=6.05 V[/tex]
