Question

Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current ????1 through R1 and the potential difference V2 across R2 .

Answer 1

Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

Answer 2

The current is 2.33 A

And, the potential difference is 15.075 V

Calculation of the current and the potential difference:

Since Two resistors, R1=3.85 Ω and R2=6.47 Ω

So,

The total resistance (Rt) is

= R1+R2

= 3.85+6.47

= 10.32 ohms.

Now here we applied ohm law

V = IR(t)..........equation 1

Here V = Emf of the battery,

I = current

R(t) = combined resistance

Now

I = V/R(t)............. Equation 2

Here

Given: V = 24 V, R(t) = 10.32 ohms

So, I should be

I = 24/10.32

I = 2.33 A.

Now the potential difference is

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

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