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

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