Question

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations

of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 32 Ω ?​

Answer 1

Answer:

lowest=2, highest=56

Explanation:

To get highest resistance we have to connect the resistors in series.

we know that

R(ser)=R1+R2+R3+R4-------------(1)

GIVEN:- R1=4 R2=8 R3=12 R4=32

PUTTING IN EQUATION (1)

R(ser)= 4+8+12+32

=56 ohm

To get lowest resistance we have to connect resisitors in parralel.

we know that

1/R(par)=1/R1+1/R2+1/R3+1/R4

PUTTING THE VALUES

1/R(par)=1/4+1/8+1/12+1/32

=24+12+8+3/96

1/R(par)=47/96

R(par)=96/47

=2 ohm(approx)