Question

In the circuit given below the potential difference across the arm PQ will be

Answer 1

Given L

R1=6ohms

R2=6ohms

R3=5ohms

V=24V

We an make out that R1,R2 and R3 are in parallel connection.

Effective combined resistance of R1 and R2:

1/Reff=1/R1+1/R2

=1/6+1/6

=2/6

Reff=6/2=3ohms

Hence total resistance of circuit is R3+ 3 ohms=5+3=8ohms

So current flowing in the circuit is :

By ohm's law= V=IR

I=V/R

=24/8=3A

The potential drop across PQ will be the same as the potential difference across the parallel combination.

So Potential drop is V= IR

=3x(/6x6/12)

=9V