An electrical source with internal reistance r is used to operate a lamp of resistance R. What fraction of the total power is delivered to the lamp?(a)R+r/R(b)R-r/R (c)R/R+r (d)R/r
Total resistance is R+r, so
R/(R+r) of the power is delivered to the load.
A good worksheet on this is to be found at
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/autbat.html#c2
Ohm my goodness, this sounds like a bright question! Let's shed some light on it!
To find the fraction of the total power delivered to the lamp, we would divide the power delivered to the lamp by the total power.
The power delivered to the lamp can be calculated using the formula P = V^2 / R, where V is the voltage across the lamp and R is the resistance of the lamp.
The total power can be calculated using the formula P_total = V^2 / (R + r), where r is the internal resistance of the electrical source.
Now let's brighten up our knowledge and simplify the formulas:
Fraction of power delivered to the lamp = P_lamp / P_total
= (V^2 / R) / (V^2 / (R + r))
= (V^2 / R) * ((R + r) / V^2)
Thankfully, the V^2 term cancels out, making life easier:
Fraction of power delivered to the lamp = R / (R + r)
So, the answer is (c) R / (R + r)!
I hope that illuminated your understanding of the topic.
To find the fraction of the total power delivered to the lamp, we need to analyze the power dissipated by the lamp and the internal resistance of the electrical source.
Let's denote the power dissipated by the lamp as P_lamp and the total power as P_total.
We know that power is given by the formula:
P = I^2 * R
where I is the current flowing through the circuit and R is the resistance.
For the lamp, the power dissipated is:
P_lamp = I^2 * R
The current flowing through the circuit is given by Ohm's Law:
I = V / (R + r)
where V is the voltage provided by the electrical source.
Substituting this expression for I into the power formula for the lamp, we get:
P_lamp = (V / (R + r))^2 * R
Now, the total power provided by the electrical source is the sum of the power dissipated by the lamp and the power dissipated by the internal resistance:
P_total = P_lamp + P_internal
P_total = (V / (R + r))^2 * R + (V / (R + r))^2 * r
Factoring out (V / (R + r))^2 from both terms, we have:
P_total = (V / (R + r))^2 * (R + r)
Now, the fraction of the total power delivered to the lamp is given by:
(P_lamp / P_total) = [(V / (R + r))^2 * R] / [(V / (R + r))^2 * (R + r)]
Simplifying the expression, we get:
(P_lamp / P_total) = R / (R + r)
Therefore, the fraction of the total power delivered to the lamp is R / (R + r).
So, the correct answer is (c) R / (R + r).
To determine the fraction of total power delivered to the lamp, we can use the concept of power in a circuit.
The power delivered to a device in a circuit can be calculated using the formula P = V^2/R, where P is power, V is voltage, and R is resistance.
Let's assume that the electrical source provides a voltage V.
The power delivered to the lamp can be calculated as PL = V^2/RL, where RL is the resistance of the lamp.
However, since the electrical source has its own internal resistance (r), the voltage across the lamp (VL) will be less than the voltage provided by the source. This is due to a voltage drop across the internal resistance of the source.
We can calculate the voltage across the lamp using Ohm's law: V = VL + Vr, where Vr is the voltage drop across the internal resistance of the source.
Applying Ohm's law to the internal resistance, we get Vr = Ir, where I is the current flowing in the circuit.
Now, we can substitute the values of Vr and V into the power formula for the lamp:
PL = (VL + Vr)^2/RL
Since Vr = Ir and Ohm's law (V = IR) can be applied to the entire circuit, we have V = (R + r)I.
By substituting the values, we get:
PL = ((RL + rL)/RL)^2*(V/rL)^2
Simplifying, we get:
PL = (RL + rL)^2/(RL)^2 * V^2/rL^2
PL = (RL^2 + 2RLrL + rL^2)/(RL)^2 * V^2/rL^2
Cancelling out the RL^2 and rL^2 terms, we get:
PL = (1 + 2rL/RL + rL^2/RL^2) * V^2/rL^2
The total power provided by the source (PS) can be calculated using the power formula:
PS = V^2/RS, where RS is the total resistance of the circuit, given by RS = RL + rL.
PS = V^2/(RL + rL)
To find the fraction of power delivered to the lamp, we divide the power delivered to the lamp (PL) by the total power provided by the source (PS):
Fraction of power delivered to the lamp = PL/PS
Fraction of power delivered to the lamp = [(1 + 2rL/RL + rL^2/RL^2) * V^2/rL^2] / [V^2/(RL + rL)]
Simplifying, we get:
Fraction of power delivered to the lamp = (RL + rL)^2/RL^2
Fraction of power delivered to the lamp = (R + r)^2/R^2
Therefore, the answer is (a) R + r / R.