Question

A battery with a terminal voltage of 9 V is connected to a circuit consisting of four 20Ω and one 10Ω resistors all in series (Figure below). Assume the battery has negligible internal resistance. (a) Calculate the equivalent resistance of the circuit. (b) Calculate the current through each resistor. (c) Calculate the potential drop across each resistor.

Answer 1

Given: Four Resistors is 20 ohms and 1 Resistor is 10 ohms, are connected in series to a terminal voltage of 9V.

a) Equivalent Resistance:

R(eq.) = R1 + R2 + R3 + R4 + R5

R(eq.) = 20 + 20 + 20 + 20 + 10

R(eq.) = 90 ohms

b) Current in each resistors:

I = V/R

I(in R1,R2,R3,R4) = 9/20 = 0.45 A

I(in R5) = 9/10 = 0.9 A

Total current supplied = 0.45A + 0.9A = 1.35 A

c) Potential drop:

E = IR

  = (1.35)(90)

E = 121.5 V

d) Power

P = IV

  = (0.9)(121.5)

P = 109.35 W