Question

Two resistors are connected in parallel and third resistor be
connected in series with the combination of parallel resistors. If this
combination be connected with a battery of the negligible internal
resistance , find the potential difference across each resistor .

Answer 1

Answer:

Given that 2Ω and 4Ω are connected in parallel,

So,

2

1

+

4

1

=

R

eq1

1

⇒

R

eq1

1

=

8

6

=

4

3

⇒

3

4

=1.3333Ω

Again 3Ω and 6Ω are coonected in parallel,

So,

3

1

+

6

1

=

R

eq2

1

⇒

R

eq2

1

=

18

9

⇒

18

9

=2Ω

Now, R

eq1

and R

eq2

are coonected in series,

So, R

eq

=R

eq1

+R

eq2

+InternalResistance

⇒R

eq

=1.333+2+0.7

⇒R

eq

=4.0333Ω

So,

I=

R

V

=

4.033

5

=1.239Amp≃1.24Amp

I

4

=

R

3

+R

2

R

3

⋅I

⇒I=

3+6

3×1.24

=

9

3.72

=0.4133Amp

So, The current through 6Ω is 0.4133Amp