Three resistors, 12 ohms, 6 ohms, and 4 ohms are connected in parallel in a clos

Question

Three resistors, 12 ohms, 6 ohms, and 4 ohms are connected in parallel in a closed circuit with a 6.0V battery of negligible resistance. Based on the above information find the following:

a) the total resistance of this circuit.

b) the current through the 12 ohm resistor.

c) the voltage drop across the 6 ohm resistor.

d) the power dissipated across the 12 ohm resistor.

e) if instead the resistors were connected in series to the battery, what would be the power dissipated across the 12 ohm resistor?

Explanation / Answer

Given :

R1 = 12 Ohms

R2 = 6 Ohms

R3 = 4 Ohms

V = 6 V

A) Total resistance = R1 || R2 || R3

=> 12 || 6 || 4

=> 1/Rtotal = 1/R1 + 1/R2 + 1/R3

=> 1/12 + 1/6 + 1/4

=> (1 + 2 + 3)/12

=> 6/12

=> 1/2

Hence Rtotal = 2 Ohms

b) Current through the 12 Ohm resistor.

Itotal = V/Rtotal

Itotal = 6 2 = 3 Amps.

Hence

I12 = Itotal(Rtotal/ R12)

I12 = 3 (2/12)

I12 = 0.5 Amps

C) Voltage drop across 6 Ohm Resistor.

Voltage drop in parallel remains the same, hence its 6V

D) Power = VI

=> 6 * 0.5

=> 3 Watts

E) If the resistors were connected in series, then the volage across the 12 Ohms resistor would be :

V12 = Vtotal(R12/Rtotal)

V12 = 6(12/ 22)

V12 = 3.27 V

The current in the series circuit remains the same, hence

I = V/Rtotal

I = 6/22

I = 0.27 Amps

Hence the power dissipation is :

P = VI

=> 3.27 * 0.27

= > 0.8829 Watts

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