Question

101. You are given several identical resistances each of value R= 10 12 and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of 5 2 which can carry a current of 4 ampere. The minimum number of resistances of the type R that will be required for this job is Yat 4 (b) 10 (c) 8 (d) 20​

Answer 1

Correct option is

C

8

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.