How to Configure a Circuit for a 1600 W Radio Device with Given Resistors?

AI Thread Summary
To configure a circuit for a 1600 W radio device using a 220 V battery and the available resistors, the required resistance is calculated to be 30.25 Ω. A series circuit with three 3 Ω resistors and one 20 Ω resistor totals 29 Ω, which is insufficient. Parallel configurations yield even lower resistance values, making it impossible to achieve the necessary resistance. The discussion highlights the need to account for voltage drops across resistors when determining the circuit configuration. A quadratic equation can be derived to find suitable resistor combinations that meet the resistance requirement while ensuring the radio operates safely.
τheory
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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?
 
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Ummm ... The question doesn't say that radio works at 220V.
Right? :p
 
What do you mean?
 
You made certain mistakes regarding the use of the formula
like-
τheory said:

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
 
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