A single resistor is wired to a battery.
Define the total power dissipated by this circuit as P(initial). What is the Power?
Now, a second identical resistor is wired in series with the first resistor. What is the power, in terms of P(initial), dissipated by this circuit?
The terms have to be in P(initial). No equations and no V or R is part of the answer.
1/2 P(intitial)
Current is reduced to half. Power is reduced by one half.
pinitial/2
Well, the power dissipated by a circuit can be defined as P = IV, where I is the current flowing through the circuit and V is the voltage across it. Since you specifically mentioned no equations, V, or R, let's get creative and keep things light!
For the initial circuit with a single resistor wired to a battery, let's call it "Bob," the power dissipated can be humorously defined as the energy Bob uses to entertain the electrons passing through him. We'll call it P(initial).
Now, when we add a second identical resistor in series with Bob, it's like turning the initially calm circuit into a double act, let's say "Bob and Tim." Now, we have more electrons to entertain and more power dissipated in the circuit.
So, the power dissipated by the circuit with the second resistor can be humorously defined as P(initial) + P(extra), where P(extra) represents the additional power required to keep Bob and Tim, the dynamic duo of resistors, performing their electrifying acts together.
To answer this question in terms of P(initial), we need to understand the relationship between power, resistance, and current.
In the given circuit with a single resistor wired to a battery, the power dissipated (P(initial)) can be determined using Ohm's Law and the formula for power:
P = I^2 * R
where P is the power dissipated, I is the current flowing through the circuit, and R is the resistance of the circuit.
Since the circuit has only one resistor and a known power P(initial), we can rearrange the formula to solve for current (I):
I = √(P(initial) / R)
Now, let's consider the second scenario where an identical resistor is wired in series with the first resistor. When resistors are in series, the total resistance (R_total) is the sum of the individual resistances:
R_total = R1 + R2
Since the resistors are identical, we can simplify this to:
R_total = 2R
To find the power dissipated by this circuit (P_final), we'll use the same formula as before, but with the total resistance:
P_final = I^2 * R_total
Substituting the expression for current (I) from the first circuit, we have:
P_final = (√(P(initial) / R))^2 * 2R
Simplifying further, we get:
P_final = (P(initial) / R) * 2R
The R values cancel out:
P_final = P(initial) * 2
Therefore, the power dissipated by the circuit with the second identical resistor in terms of P(initial) is twice the initial power (2 * P(initial)).