Question

Three resistors having resistances of 1.60 Omega, 2.40 Omega and 4.80 Omega are connected in parallel to a 28.0 V battery that has negligible internal resistance. Find (a) the equivalent resistance of the combination. (b) the current in each resistor. (c) the total current through the battery. (d) the voltage across each resistor. (e) the power dissipated in each resistor. (f) which resistor dissipates the maximum power the one with the greatest resistance or the least resistance? Explain why this should be.

Answer 1

Answer:

(a)

Equivalent resistance is

R

eq

=R

1

+R

2

+R

3

=(1+2+3)Ω=6Ω

(b)

Current in the circuit is given by I=

R

E

I=

6

12

=2Ω

Potential drop across 1Ω, V

1

=IR

1

=2×1=2V

Potential drop across 2Ω,V

2

=IR

2

=2×2=4V

Potential drop across 3Ω,V

3

=IR

3

=2×3=6V