The initial power is $$P = V^2/R_1$$. When $$R_2$$ is added in series, the new total resistance is $$R_{\text{series}} = R_1 + R_2$$. Since $$R_1+R_2 > R_1$$, the new total resistance is greater than the original resistance. The new total power is $$P_{\text{new}} = V^2/R_{\text{series}}$$. Because the denominator has increased, the new total power is less than the original power $$P$$.

First, find the resistance of the wire using the formula $$R = \frac{\rho L}{A}$$, where the cross-sectional area $$A = \pi r^2$$. So, $$R = \frac{\rho L}{\pi r^2}$$. Then, use the power formula $$P = \frac{V^2}{R}$$. Substituting the expression for R gives $$P = \frac{V^2}{\rho L / (\pi r^2)} = \frac{V^2 \pi r^2}{\rho L}$$.

The claim is incorrect. Let the battery voltage be $$V$$. The power dissipated by $$R_1$$ alone is $$P_1 = V^2/R_1$$, and by $$R_2$$ alone is $$P_2 = V^2/R_2$$. Their sum is $$V^2(1/R_1 + 1/R_2)$$. In series, the total resistance is $$R_s = R_1 + R_2$$, and the total power is $$P_s = V^2/(R_1 + R_2)$$. These expressions are not equal. The physical reason is that the current drawn from the battery changes. In the series circuit, the current is $$I_s = V/(R_1+R_2)$$, which is smaller than the current through either resistor when connected alone.

The power dissipated in the external resistor is $$P = I^2 R = \left(\frac{\mathcal{E}}{R+r}\right)^2 R$$. Since $$\mathcal{E}$$ and $$R$$ are the same for both cases, power depends on the total resistance $$R+r$$. As $$r_B < r_A$$, the total resistance of the circuit with Battery B, $$R+r_B$$, is less than that with Battery A, $$R+r_A$$. A smaller total resistance results in a larger current. Since power is proportional to the square of the current, $$P_B > P_A$$.

By the principle of conservation of energy, the total energy dissipated by the resistor must be equal to the total energy initially stored in the capacitor. The initial energy stored in the capacitor is given by $$U_C = \frac{1}{2} C V_0^2$$. This entire amount of energy is converted into thermal energy in the resistor as the capacitor discharges to zero potential.

The power dissipated by a resistor is given by $$P = IV = I^2R = V^2/R$$. Since the resistors are not identical and the currents and voltages are generally different for each, knowing only the current (A), voltage (B), or resistance (C) is insufficient. For example, a large current in a small resistor may dissipate less power than a smaller current in a large resistor. One must calculate the power for each resistor using one of the formulas, which requires knowing $$I$$ and $$V$$, or $$I$$ and $$R$$, or $$V$$ and $$R$$. The product $$IV$$ is the definition of power and is always sufficient.

According to the maximum power transfer theorem, a source with internal resistance $$r$$ delivers maximum power to an external load resistor $$R$$ when $$R=r$$. This can be proven by taking the derivative of the power expression $$P = I^2 R = (\frac{\mathcal{E}}{R+r})^2 R$$ with respect to $$R$$ and setting it to zero.

This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an 18.0 V battery with total current I=3.0 A, requiring calculation of power supplied by the battery. Choice B (54 W) is correct because it applies the formula P=IV using values I=3.0 A and V=18.0 V, resulting in P=(3.0 A)(18.0 V)=54 W. Choice D (54 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that battery power output is always P=IV where I is total current drawn and V is battery voltage. Practice with parallel circuits to understand how total current relates to individual branch currents.

This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes two resistors in series with R₂=6.0 Ω and current I=1.0 A, requiring calculation of power dissipated by R₂. Choice A (6.0 W) is correct because it applies the formula P=I²R using values I=1.0 A and R=6.0 Ω, resulting in P=(1.0 A)²(6.0 Ω)=6.0 W. Choice D (6.0 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize unit consistency and that power dissipation in a resistor is always P=I²R when current is known. Practice recognizing series circuits where the same current flows through all components.

This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an extension cord with R=0.20 Ω carrying I=8.0 A, requiring calculation of resistive power loss. Choice B (12.8 W) is correct because it applies the formula P=I²R using values I=8.0 A and R=0.20 Ω, resulting in P=(8.0 A)²(0.20 Ω)=64×0.20=12.8 W. Choice D (12.8 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that power loss in wires is continuous dissipation measured in watts. Practice calculating wire losses to understand why thick wires are used for high currents.