physics help?

Question

When two identical resistors are connected in series across the terminals of a battery, the power delivered by the battery is 20.0 W. If these resistors are connected in parallel across the terminals of the same battery, what is the power delivered by the battery?

I really need help, i don't know how to do this, answers with explanations pls, best answer gets 10 pts.

Answer 1

The power dissipated by a resistor is P = V^2 / R. In this case, let us call the resistance of each resistor R and the voltage of the battery V. Then, in the first situation (series resistance), the total resistance in the circuit is 2R, and the power is V^2 / 2R. In the second situation (parallel resistance), the total resistance is R / 2, so the power is V^2 / (R / 2) = 2V^2 / R. This is 4 times the power of the first situation, so the power would be 4*(20.0 W) = 80.0 W.

Note that for series resistance, you just add the resistors. For parallel resistance, with resistors R1 and R2, the equivalent resistance is 1 / ((1 / R1) + (1 / R2)). But for n many equal resistors R in parallel, the equivalent resistance reduces to R / n. In this case, n was 2.

Answer 2

Not really physics - more like algebra.

You are being asked to manipulate the two basic equations govening electricity.

R=E/I
(Resistance is the ration between (E) Voltage & (I) Current

&

P=E*I (Power equals Voltage * Current)

You also need to know that Voltage (E) is like pressure or force (E is short for Electromotive Force) and Current (I) is like flow. So P=E*I is like saying the power you can get from a hose, is the water pressure (E) times the flow rate(I).

The rest is simple math. The flow through series resisters must be the same. (Garden hoses hooked together have the same water flowing through them - Right )

Since the flow is identical, & the Ration to Pressure (E) to flow (aka resistance) is the same. Both resisters must have the same indivual Pressure (E) across them which must be half the total pressure(voltage) provided by the battery.

When we swich to the parallel configuration, the voltage (pressure) across each resister is now double what is was in the serial configuration. Since I = E/R this means the flow through each resistor has doubled also. (Right - double the pressure, double the flow)

And since we've double both the pressure & flow, the power throuh each resistor must have quadrupled because Power=E*I. (So: 2E*2I = 4P)

In series each resistor was burning half of the total power i.e. 10 Watts, which means in parallel each is burning 4 times that. I.e. 40 watts a piece for a total of 80 watts.

Hope this helps.

Answer 3

Series resistance is R1 + R2
parallel resistance is R1*R2/(R1+R2)

if R1=R2 then series is 2R
parallel is R^2/2R = R/2

so the series resistance (2R) is four times the parallel resistance (R/2)

so if we let the parallel resistance be X, series is 4X

Power(W) is current(I)*voltage(V), (i) W=IV
and V=IR (rewrite as (ii) I=V/R)
substituting (ii) into (i) W=V^2/R
a battery is assumed to produce constant voltage so V^2 is a constant (K) so W=K/R

The unknown wattage is Z

substitute our resistance X for the variable R
we know (1) 20=K/4X and (2) Z=K/X
substitute (2) in (1) 20=Z/4 Z=80

very long winded but shows all steps :)

Answer 4

Let R = resistance of the resistor in Ω.
When connected in series, equivalent resistance = 2R.
When connected in parallel, equivalent resistance is R/2.

Power, P = V^2 / R.
=> Power is inversely proportional to rsistance.
=> 20 = v^2 / 2R
and P = v^2 / 0.5R
=> P/20 = 2R / 0.5R = 4
=> P = 80 W.

Answer 5

Let resistance be R.
Potential diffrence be V

Power = V²/R

When they are connected in series combined Resistance
2R
20=V²/2R
V²=40R

Now when they are connected in parallel , combined resistance = R/2

Power=V²/R
R=R/2
V²=40R

Power = 40R/R/2
Power=80W