AP EAPCET 04-Jul-2022 Shift 2 Paper

The circuit diagram is shown as,
Applying KVL in loop 3, we get
I3R+R(I3−I2)=0
2I3R−RI2‌=0
I3‌=‌

I2

2

.......(i)
Applying KVL in loop 2, we get
‌RI2−3ε+R(I2−I1)+2ε+RI2−R(I2−I3)=0
‌⇒‌‌2RI2−RI1+RI3−ε=0
‌⇒2RI2−RI1+R‌

I2

2

−ε=0
‌⇒‌‌5RI2−2RI1−2ε=0.........(ii)
By applying KVL in loop 1, we get
‌RI1+ε−2ε+R(I1−I2)=0
2RI1−RI2−ε=0
I=‌

RI2+ε

2R

.......(iii)
From eqs(ii) and (iii), we get
5RI2−2R‌

(RI2+ε)

2R

−2ε=0
5RI2−RI2−ε−2ε‌=0
⇒‌‌I2‌=‌

3ε

4R