Consider the circuit shows below. Take the emf of the battery to be V_0 = 5V and

Question

Consider the circuit shows below. Take the emf of the battery to be V_0 = 5V and R = 10 ohm. Assume that the wires are perfect conductors and that the battery has no internal resistance Find the current in the circuit Find the power generated by the battery Find the power dissipated in the resistor. Consider again the circuit of problem 3. Again take the emf of the battery to be V_0 = 5 V and R = 10 ohm. But in this problem take the interval resistance of the battery to be 0.15 ohm and the two wire that connect the battery to the resistor to each be 14 gauge gold wire of length 1 m Find the current in the circuit Find the power generated by the battery Find the power dissipated in the resistor Find the power dissipated in the wires Find the power dissipated in the internal resistance of the battery Find the terminal voltage across the battery.

Explanation / Answer

Hi,

In this case, as there are many questions, I'm going to answer until part A of Problem 4, however by doing the first problem and considering some stuff that I will say to solve the first part of the second one, I'm sure you will be able to do the rest.

First we need to know certain relations:

I = V/Req (1) ; where I is the current, V is the voltage of the battery and Req is the equivalent resistance of the circuit (if there's only one resistance, then Req = R).

Pj = I2 Rj (2) ; where Pj is the power dissipated in the j-component, I is the current though said component while Rj is the resistance of that component.

P = I2 Req (3) ; where P is the power generated by the battery, I is the current through the circuit and Req is the equivalent resistance of the circuit.

Req = R1 + R2 + R3 + ....... + Rn (4) ; if the components are in series

1/Req = 1/R1 + 1/R2 + 1/R3 + ....... + 1/Rn (5) ; if the components are in parallel

Rw = p L/S (6); where Rw is the resistance of a wire, p is the resistivity of the material, L is the length of the wire and S is the cross sectional area of said wire.

dn = 0.127*92[ (36 - n)/39 ] (7); where d is the diameter of the wire in mm and n is the gauge number.

V = Vo - rI (8); where V is the terminal voltage of the battery, Vo is the voltage of the battery, I is the current thought the circuit and r is the internal resistance.

Problem 3.

In this circuit the only resistance is the one in the resistor, so in all previous equations, Req = R and the power generated by the battery is the same that is dissipated in the resistor.

A) Using (1) we solve this part:

I = V/R = 5 V / 10 = 0.5 A

B) Using (3) this can be solved:

P = I2 R = (0.5 A)2 (10 ) = 2.5 W

C) As it was explained before, this value is equal to the one above, so:

P = 2.5 W

Problem 4.

In this case there's more than one resistance but all of them are in series. So, we have to find out the other resistances and calculate the equivalent:

Using a table we find out that gold's resistivity is equal to :   p = 2.44*10-8 m

Through (6) and (7) we find the resistance of the wires:

d14 = 0.127*92[22/39] = 1.63 mm = 1.63*10-3 m

Rw = (2.44*10-8 m)*(1 m / [(/4)*(1.63*10-3 m)2] ) = 0.012

So, we can finally use (4) to find the equivalent resistance:

Req = (10 + 2*0.012 + 0.15) = 10.174

A) Again, using (1) we have the following:

I = 5 V / 10.174 = 0.491 A

B) Is done with (3)

C) , D) and E) are done using (2), considering the resistances of each component (wires, resistor, and the battery).

F) Is done with (8)

If you still have doubts, post the rest as an independent question.

I hope it helps.

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