Current Electricity Part 3

Method-I:
‌Req=2+‌

5R

5+R

=‌

10+7R

5+R

‌i=‌

3

Req

=‌

3(5+R)

10+7R

‌i1=‌

2

5

,i2=‌

2

R

‌i=i1+i2
‌‌

3(5+R)

10+7R

=‌

2

5

+‌

2

R

=‌

2(5+R)

5R

‌15R(5+R)=2(5+R)(10+7R)
‌75R+15R2=2(50+35R+10R+2R2)
‌15R2+75R=14R2+90R+100
‌R2−15R−100=0
‌R=‌

15√225×1×100

2

‌=‌

15±√625

2

=‌

15±25

2

‌R=20Ω
Method-II:
Given potential across 5Ω and voltmeter is 2V. To find resistance R of voltmeter.
Let current in 5Ω be i1, and in Ri2.
i1=‌

2

5

‌ and ‌i2=‌

2

R

V across 2Ω will be 1 volt and i=‌

1

2

A.
Using junction law: i=i1+i2
‌‌

1

2

=‌

2

5

+‌

2

R

‌‌

2

R

=‌

1

2

−‌

2

5

=‌

1

10

‌R=20Ω