Question

Review Part A In the circuit shown in the diagram(Figure 1), suppose R1 R2 210 and Rs R The emf of the battery is 12.0 V Find the value of R such that the current supplied by the battery is 0.0730 A Submit Incorrect; Try Again; 4 attempts remaining Part B Find the value of R that gives a potential difference of 3.65 V across resistor 2 Submit Provide Feedback Next > Figure 1 of 1 R2

Answer 1

a)

net resistance,

R = V/i = 12/ 0.073 = 164.383

R1 * ( R2 + R3) / ( R1 + R2 + R3) = 164.383

210* ( 210+ R3) / ( 2*210 + R3) = 164.383

210*210 + 210*R3 = 164.383*2*210+ 164.383*R3

R3 = 546.74 ohm

=============

b)

as we know from Ohm's law

for constant value of current V is directly proportional to resistance

V3 / V2 = R3 / R2

(12-3.65)/ 3.65 = R3 / 210

R3 = 480.41 ohm

===========

do comment in case any doubt, will reply for sure.. Goodluck