Two resistors, R 1 = 2.01 Ω R1=2.01 Ω and R 2 = 5.29 Ω R2=5.29 Ω , are connected in series to a battery with an EMF of 24.0 24.0 V and negligible internal resistance. Find the current I 1 I1 through R 1 R1 and the potential difference V 2 V2 across R 2 R2 .

I1 = 3.288 A,

V2 = 17.39 V

Explanation:

The combined resistance of the two resistor connected in series is given as

R' = R1+R2 Equation 1

Where R' = Combined resistance.

Given: R1 = 2.01 Ω, R2 = 5.29 Ω

Substitute into equation 1

R' = 2.01+5.29

R' = 7.3 Ω

Using

E = I(R'+r) Equation 2

Where E = emf of the battery, I = current through the circuit, r = internal resistance.

Given: E = 24 V, R = 7.3 Ω, r = 0 Ω( Negligible)

Substitute into equation 2

24 = I(7.3)

I = 24/7.3

I = 3.288 A.

Since the resistors are connected in series, the same amount of current flows through them

Therefore,

I = I1 = 3.288 A.

Using ohm's law,

V2 = IR2 Equation 3

Where V2 = potential difference across R2 resistor.

Given: I = 3.288 A, R2 = 5.29 Ω

Substitute into equation 3

V2 = 3.288(5.29)

V2 = 17.39 V

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