Question

2. Two resistances of 6 Q and 12 Q are connected in parallel. Their net resistance is
a) 702
b) 60
c) 40
d) 50​

Answer 1

Resistance, R1 = 6 Ω

Resistance, R2 = 12 Ω

==> 1/R = 1/R1 + 1/R2

==> 1/R = 1/6 + 1/12

==> 1/R = 2 + 1/12

==> 1/R = 3/12

==> 1/R = 1/4

==> R = 4 Ω

Hence, The net resistance is 4 Ω

Answer 2

Answer:

It would be 4 ohms

Explanation:

Q is unit of charge not resistance

here to calc. net resistance, we have formula (r1×r2)/r1+r2

=6×12/6+16=72/18

=4 Ohm