Question

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

Answer 1

(a) The equivalent resistance of three resistors in series is the sum of the three resistances:

`R_(eq)=R_1 + R_2 + R_3 = 37 \Omega + 17 \Omega + 110 \Omega = 164 \Omega`
And so the current flowing through the circuit (and the battery) can be found by using Ohm's law:

`I= (V)/(R_(eq))= (9.0 V)/(164 \Omega)=0.055 A`

(b) The potential difference across each resistor is given by Ohm's law:

`V_(37) = I R_(37)=(0.055 A)(37 \Omega)=2.04 V`

`V_(17)=I R_(17)=(0.055 A)(17 \Omega)=0.94 V`

`V_(110)=I R_(110)=(0.055 A)(110 \Omega)=6.05 V`