Question

f
8. You are given several identical resistors each
of value 512 and each capable of carrying a
maximum current of 2A. It is required to make a
suitable combination of these resistances to produce
a resistance of 2.5, which can carry current of 4A.
The minimum number of resistances required for
this job is
1) 2. 2) 4 3) 6 4) 8​

Answer 1

Resistance of each resistor R=10

Equivalent resistance of each segment R

′

=R+R=2R

∴ R

′

=2×10=20

Such 4 segments are connected in parallel to each other.

Thus equivalent resistance of the circuit R

p

​

=

4

R

′

​

⟹ R

p

​

=

4

20

​

=5

Maximum current flowing through each resistor is one ampere i.e. i=1A

Thus total current flowing through the circuit I=i+i+i+i

∴ I=4i=4A

Thus minimum 8 resistors are required to satisfy the above conditions.

Answer 2

Answer:

THE CORRECT ANSWER IS 2