Question

in a circuit three 3Ω resistances are connected in parallel and this combination is connected in series with another 3Ω resistance. What is the equivalent resistance of the
Plz. do it fast

Answer 1

Given: A resistance of 3Ω is connected in series with another resistance of 5Ω. A potential difference of 6V is applied across the combination.

To find the current through the circuit and potential difference across the 3Ω resistance.

Solution:

as the resistance are connected in series

so R=R

1

+R

2

=3+5=8Ω

now according to Ohm's law,

R=IR

⟹I=

R

V

⟹I=

8

6

⟹I=0.75A

So, 0.75A of current flow through circuit.

Now , Potential Difference across the 3Ω resistance is

V

1

=IR

1

⟹V

1

=0.75×3

⟹V

1

=2.25V

Now the corresponding current will be,

I

1

=

R

1

V

1⟹I

1

=

3

2.25

⟹I

1

=0.75A=I

and we know that the same current flow through each of the resistance in series

Answer 2

Answer:

ANSWER

Suppose R is the total resistance of parallel combination then it is given by

R

1

=

R

1

1

+

R

2

1

+

R

3

1

Given - R

1

=R

2

=R

3

=3Ω

or

R

1

=

3

1

+

3

1

+

3

1

=1

Thus R=1 Ohm.