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Complete question is missing, so i have attached it.
Answer:
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

(A) The current when connected across ab is 3.5A
(B) The current when connected across bc is 4.5 A
(C) The current when connected across ac is 3.15A
(D) The current when connected across bc with internal resistance is 3.25 A
Equivalent resistance and current:
(a) When connected across ab, the resistance between ac and bc is;
20Ω + 10Ω = 30Ω
It is parallel with 15Ω resistor.
Thus, equivalent resistance is;
R = (30 × 15)/(30 + 15)
R = 450/45
R = 10Ω
Now current is given by
I = (V/R)
I = 35/10
I = 3.5A
(b) When connected across bc, the resistance between ab and ac is;
15Ω + 20Ω = 35Ω
It is in parallel combination with the 10Ω resistor
Thus, equivalent resistance is;
R = (35 × 10)/(35 + 10)
R = 350/45
R = 7.778Ω
Now the current is:
I = 35/7.778
I = 4.5 A
(c) When connected across ac, the resistance between ab and bc is;
15Ω + 10Ω = 25Ω
it is now parallel with the 20Ω resistor.
Thus, - is;
R = (25 × 20)/(25 + 20)
R = 11.11Ω
Current is;
I = 35/11.11
I = 3.15A
(d) when connected across bc, the internal resistance will be added to the circuit resistance across bc.
So, equivalent resistance = 7.778 + 3 = 10.778Ω
Current is;
I = 35/10.778
I = 3.25 A
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