Question

1. for the circuit in figure 5 - 1, assume a student accidentally set the power supply to 24 v instead of 12 v.

(a) how much power would be dissipated in the resistor? ____________
(b) would a 1/4 w resistor be adequate for this case? ____________

1. for the circuit in figure 5 - 2, what was happening to the total power in the circuit as the resistance of r₂ was increasing? explain your answer.

Explanation:

Step1: Recall power - voltage - resistance formula

The power dissipated in a resistor is given by $P=\frac{V^{2}}{R}$. Since the resistance value is not given, assume the resistance $R$ is constant. The original voltage $V_1 = 12\ V$ and the new voltage $V_2=24\ V$.

Step2: Calculate power for new voltage

Let the power for $V_1$ be $P_1=\frac{V_1^{2}}{R}$ and for $V_2$ be $P_2=\frac{V_2^{2}}{R}$. Substituting $V_2 = 24\ V$ into the power formula $P=\frac{V^{2}}{R}$, if we assume the original power $P_1$ with $V_1 = 12\ V$ and new power $P_2$ with $V_2=24\ V$, and since $P_1=\frac{12^{2}}{R}=\frac{144}{R}$ and $P_2=\frac{24^{2}}{R}=\frac{576}{R}$. If we assume $R = 1\ \Omega$ (for the sake of calculation, since the ratio will be the same for any non - zero $R$), $P_2=\frac{24^{2}}{1}=576\ W$. In general terms, since $P=\frac{V^{2}}{R}$, when $V$ doubles from $12\ V$ to $24\ V$, $P$ becomes 4 times the original value.

Step3: Check resistor adequacy

We have found that the power dissipated is much larger. A $\frac{1}{4}\ W$ resistor is not adequate because the power dissipated ($P_2$) is likely to be much greater than $\frac{1}{4}\ W$.

Step4: Analyze total power as $R_2$ increases

The total resistance in a circuit (assuming a simple series or parallel circuit) changes as $R_2$ increases. For a series circuit $R_{total}=R_1 + R_2+\cdots$, and for a parallel circuit $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots$. The total power in a circuit is $P=\frac{V^{2}}{R_{total}}$. As $R_2$ increases, $R_{total}$ increases (in series) or the equivalent resistance in parallel also changes in a way that $R_{total}$ increases. Since $P=\frac{V^{2}}{R_{total}}$ and $V$ is constant, as $R_{total}$ increases, the total power $P$ decreases.

Answer:

(a) 4 times the power at 12 V (or if $R = 1\ \Omega$, $576\ W$)
(b) No
For question 3: The total power in the circuit is decreasing as the resistance of $R_2$ is increasing because $P=\frac{V^{2}}{R_{total}}$ and $R_{total}$ increases as $R_2$ increases while $V$ is constant.