Question

Three resistors with resistances of 10 Ω, 6.67 Ω and 4.25 Ω are connected in parallel to each other and to a 12V battery. What is the total current in this circuit?

Answer 1

As given,

Reistances, 10Ω , 6.67Ω & 4.25 Ω are in parallel.

The equivalent resistance in parallel is given by formula:

1/R(eq) = 1/R1 + 1/R2 +1/R3 ...

So, putting values,

1/R(eq) = 1/10 + 1/6.67 + 1/4.25 = 0.485

R(eq) = 1/0.485 = 2.06 Ω

As given this R(eq) is connected to 12 V battery in parallel hence the toatal current in the circuit is given by ohm's law,

I = V/R

Putting, V = 12V, R = R(eq) = 2.06Ω

I = 12/2.06 = 5.82 A

Hence the toatal current in the circuit is = 5.82 A.

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