Two resistors, of r1 = 3.11 ω and r2 = 6.15 ω, are connected in series to a battery with an emf of 24.0 v and negligible internal resistance. find the current i1 through the first resistor and the potential difference δv2 between the ends of the second resistor.

Part (a):
1- Since the resistors are in series, therefore, the total resistance is the summation of the two resistors.
Therefore:
Rtotal = R1 + R2 = 3.11 + 6.15 = 9.26 ohm

2- Since the two resistors are in series, therefore, the current flowing in both is the same. We will use ohm's law to get the current as follows:
V = I*R
V is the voltage of the battery = 24 v
I is the current we want to get
R is the total resistance = 9.26 ohm
Therefore:
24 = 9.26*I
I = 24 / 9.26
I = 2.59 ampere

Part (b):
To get the voltage across the second resistor, we will again use Ohm's law as follows:
V = I*R
V is the voltage we want to get
I is the current in the second resistor = 2.59 ampere
R is the value of the second resistor = 6.15 ohm
Therefore:
V = I*R
V = 2.59 * 6.15
V = 15.9285 volts

Hope this helps :)

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