Question

What is (a) highest, and (b) lowest, resistance which can be obtained by combining four resistors having the following resistances?; 4Ω, 8Ω, 12Ω, 24Ω

Answer 1

Answer:

(a) Highest resistance is achieved by a series combination :

R = R1 + R2 + R3 + R4

= 4Ω + 8Ω + 12Ω + 24Ω

=  48Ω

In series resistor combination, the resistors are placed in series in the circuit. If one resistor is removed then the current stops flowing through the whole circuit. A series combination always gives the highest equivalent resistance of the circuit.

 

(b) Lowest resistance is achieved by parallel combination:

1/R = 1/R1 + 1/R2 + 1/R3 +1/R4

= 1/4 + 1/8 + 1/12 + 1/24

= 1/2

R = 2Ω

In parallel resistor combination, the resistors are placed in parallel order in the circuit. One terminal of all resistor is connected together and another terminal of all resistors are connected together. If one resistor is removed then the current will still flow through the other resistors. A parallel combination always gives the lowest equivalent resistance of the circuit.

Answer 2

Answer:

(a) If these coils are connected in series, then the equivalent resistance will

be the highest, as R=R1+R2+R3+R4

∴ R=4+8+12+24 = 24Ω

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by

⇒R=2 Ω

∴ 2 Ω is the lowest total resistance.

Explanation: