Question

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 .

Answer 1

Answer:

I = 3.3 A

V2 = 17.4 V

Explanation:

a)

• In a circuit series, the current is the same through any point of the circuit.
• Assuming the resistors are in the linear zone of operation, we can apply Ohm's Law to both resistors, as follows:

        $V = I* r_{eq} = V_{R1} + V_{R2} = (I*R_{1}) + (I*R_{2}) = I * (R_{1} + R_{2})\\ \\ R_{eq} = R_{1} + R_{2}$

• Therefore, we can find the current I as follows:

        $I =\frac{V}{R_{eq} } = \frac{24.0 V}{7.3 \Omega} = 3.3 A$

b)

• Applying Ohm's law to R2, we can find the voltage through R2 as follows:

       $V_{R2} = I* R_{2} = 3.3 A * 5.29 \Omega = 17.4 V$

Answer 2

Answer:

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