Current Electricity Part 2

According to given circuit diagram, As we know that, parallel equivalent resistance,

1

Req

=1R1+

1

R2

+

1

R3

+...
and series equivalent resistance,
Req=R1+R2+R3+...
Let the net resistance across a and b be R’
∴

1

R′

=

1

2+2+6

+

1

4+6

a=

1

10

+

1

10

=

2

10

R' = 5 Ω
Hence, total resistance, R=R′+1=5+1=6Ω
According to current division rule, current in upper branch,
I1=I.

4+6

(2+2+6)+(4+6)

=1.

10

20

=

1

2

=

1

2

.

V

R

=

1

2

×

12

6

=1A
Again, according to current division rule, current in 15Ω resistor
I15=I1.

10

10+15

=1×

2

5

=0.4A
∴ Voltage drop across15Ω resistor,
V15=I15×15=0.4×15=6V