Answer to Question #131123 in Electric Circuits for Titomi

Question

A battery of emf 24v and internal resistance r is connected to a circuit having two parallel resistors of 3 Ohms and 6 Ohms in series with an I ohm resistor. Consider the current flowing in the 3 Ohms resistor is 0.8A, calculate the current in the 6 Ohms resistor.

Answer 1

V= 24 Volts, internal resistance r

The value of the third resistance is not defined thus I take the assumptions that, the resistance of the parallel connected resistors = R₁ = R₂ = 3 Ohms

R _{(eq 12)} = (R₁ × R₂)/ (R₁ + R₂) = (3 × 3) / (3 + 3) = 1.5 Ohm

R3 = 6 Ohms

The equivalent resistance of the whole circuit will then be

R _(eq) = R _(eq12) + R₃ = 1.5 + 6 = 7.5 Ohm

Therefore, V = IR, I = (V/R) = (24/7.5) = 3.2 A

If the current in 3 Ohm resistor is 0.8 A, and they are in parallel with the 6 Ohms resistor,

Therefore, the current in the 6 Ohm resistor will be,

= 3.2 A – 0.8 A = 2.4 A