What current will this array draw from a 35.0 V battery having negligible internal resistance if we connect it across ab.

Complete question is missing, so i have attached it.

A) I = 3.5A

B) I = 4.5 A

C) I = 3.15A.

D)I = 3.25 A

Explanation:

a) When we connect across ab, the resistance from ac and bc is;

20ohms + 10ohms = 30 ohms.

This will now be parallel with the 15 ohms resistor.

Thus, equivalent resistance is;

R = (30 × 15)/(30 + 15)

R = 450/45

R = 10 ohms

Formula for Current is; I = (E/R)

Thus;

I = 35/10

I = 3.5A.

b) similarly, when we connect across bc, resistance from ac and ab is;

15 ohms + 20 ohms = 35 ohms.

This will now be parallel with the 10 ohms resistor.

Thus, equivalent resistance is;

R = (35 × 10)/(35 + 10)

R = 350/45

R = 7.778 ohms.

Current is;

I = 35/7.778

I = 4.5 A.

c) similarly, when we connect across ac, resistance from ab and bc is;

15 ohms + 10 ohms = 25 ohms.

This will now be parallel with the 20 ohms resistor.

Thus, equivalent resistance is;

R = (25 × 20)/(25 + 20)

R = 11.11 ohms.

Current is;

I = 35/11.11

I = 3.15A.

d) since connected across bc, the internal resistance will be added to the circuit resistance, which was seen to be 7.778 ohms in step b above

Thus, equivalent resistance = 7.778 + 3 ohms = 10.778

Current is;

I = 35/10.778

I = 3.25 A

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.