Construction of a Circuit

[τheory]

Homework Statement

You are given a battery of 220 V and a radio device that requires 1600 W to run. You are also given four resistors: three 3 Ω resistors and one 20 Ω resistor. What circuit configuration will allow the radio device to run without getting fried?

The Attempt at a Solution

Power = (Voltage)²/Resistance
1600 W = (220 V)²/R
R = 30.25 Ω

So this device would require 30.25 Ω to function correctly. However, based on several circuit configurations, this magnitude of resistance isn't obtainable? For example, if I used just a complete series circuit, then the total resistance would be:

R total = R1 + R2 + R3 + R4
R total = 3 + 3 + 3 + 20 = 29 Ω total which isn't exactly 30.25

Incorporating a parallel circuit would cause resistance values to be even less than their original values because parallel circuits require the reciprocal resistance values. So if I had a circuit consisting of all the resistors parallel, then I would get:

1/R total = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/R total = 1/3 + 1/3 + 1/3 + 1/20
1/R total = 0.95 Ω, which is way less

Thus, can someone point me in the right direction?

 

[cupid.callin]

Ummm ... The question doesn't say that radio works at 220V.
Right? :p

 

[τheory]

What do you mean?

 

[Zubeen]

You made certain mistakes regarding the use of the formula
like-

Power = (Voltage)²/Resistance
1600 W = (220 V)²/R
R = 30.25 Ω

Since the Resistance is to be attached to the circuit , the V=220 V will be wrong as there will be potential drop across the resistor (if attached in series to the radio).
So we must go like this,
Let V₁ be the P.D (potential drop) across the resistor combination (R₀ ), and V₂ be potential drop across the radio.
Then we have that V₁ + V₂ = 220.
so since its a series combination of net resistance of the combination and the radio, the current through both of them will be the same.
then,
we have
IR₀ + 1600/I =220
on solving , you will get a quadratic equation in terms of I.
Further, since I can be real number only so the D of quadratic equation must be >0
that will give =>
R₀ < 121/16
so any combination satisfying above inequality will be the result and similarly , you can check for the parallel combination of radio and net resistance.


Zubeen

 

[asteriatic]

I would suggest using a combination of series and parallel circuits to achieve the desired resistance of 30.25 Ω. This can be done by connecting two of the 3 Ω resistors in parallel, giving a total resistance of 1.5 Ω, and then connecting this combination in series with the remaining 3 Ω resistor and the 20 Ω resistor. This will give a total resistance of 24.5 Ω, which is close to the desired value of 30.25 Ω. This circuit configuration will allow the radio device to run without getting fried, as it provides the correct amount of resistance to limit the flow of current and prevent overheating. It is important to carefully consider the circuit configuration and the values of the resistors in order to ensure the safe and efficient operation of the radio device.