2021 A/L Physics Past Paper · Question 46 (English Medium)

Answer: (4) $6\ \text{W}, 12\ \text{W}$

Given information

We know:

The battery has e.m.f. $12\ \text{V}$ and internal resistance $2\ \Omega$. Device $X$ is $6\ \Omega$ and device $Y$ is $3\ \Omega$. They are side by side across the battery.

What each letter means:

$E$: the e.m.f. of the battery (its full push), $12\ \text{V}$.

$r$: the internal resistance of the battery (its own resistance), $2\ \Omega$.

$R_X$ and $R_Y$: the resistances of the devices, $6\ \Omega$ and $3\ \Omega$.

$R_p$ ($\Omega$): the resistance of the two devices together.

$R_T$ ($\Omega$): the total resistance.

$I$ ($\text{A}$): the current from the battery.

$V$ ($\text{V}$): the voltage across the devices (the voltage at the battery terminals).

$P_X$ and $P_Y$ ($\text{W}$): the power used by each device.

What the question asks

We want: the power used by $X$ and the power used by $Y$.

The plan: find the one resistance that stands in for both devices. Then find the current from the battery. Then find the voltage that is left for the devices. At last, use that voltage to get each power.

Relevant theory

resistance acts against the flow of the current inside the cell and it is called the internal resistance of the cell.

Given above is an expression for the potential difference between the terminals of a source when a current is flowing through the source.

Passive resistors are those resistors in which the electrical energy flowing through them gets completely converted into heat.

the total charge entering a junction of a circuit should be equal to the total charge leaving the junction.

The current flowing in a conductor is directly proportional to the potential difference between its ends provided temperature and its other physical factors remain constant

Required equations

$V = IR$

$\dfrac{1}{R_p} = \dfrac{1}{R_X} + \dfrac{1}{R_Y}$

$R_T = R_p + r$

$V = E - Ir$

$P = \dfrac{V^2}{R}$

Applying it to this question

Think of the battery as a water pump with a narrow part inside it. The narrow part is $r$. Some of the pump's push is used up there.

Two taps, $X$ and $Y$, are side by side on the pipe from the pump. The same push is on both taps. At the junction the flow splits, and the two parts add up to the whole flow.

Each tap has the same voltage $V$ across it. So $X$ carries $\dfrac{V}{R_X}$ and $Y$ carries $\dfrac{V}{R_Y}$, and the total is $I = V\left(\dfrac{1}{R_X} + \dfrac{1}{R_Y}\right)$. That is the same as one resistor $R_p$ with $\dfrac{1}{R_p} = \dfrac{1}{R_X} + \dfrac{1}{R_Y}$.

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}$

$2$

$R_p = 2$

$4$

$R_T = R_p + r = 2 + 2$

$3$

$I = \dfrac{E}{R_T} = \dfrac{12}{4}$

$6$

$Ir = 3 \times 2$

$V = E - Ir = 12 - 6$

$P_X = \dfrac{V^2}{R_X} = \dfrac{6^2}{6}$

$12$

$P_Y = \dfrac{V^2}{R_Y} = \dfrac{6^2}{3}$

$\dfrac{V^2}{R}$ has units $\dfrac{\text{V}^2}{\Omega} = \text{V} \times \dfrac{\text{V}}{\Omega} = \text{V} \times \text{A} = \text{W}$. So the answers are in watts.

Reasoning and calculation

The battery loses $6\ \text{V}$ on its own resistance, and $6\ \text{V}$ is left for the two devices. The same $6\ \text{V}$ is across both of them.

A device takes power $\dfrac{V^2}{R}$. With the same voltage, the smaller resistance takes more power. $Y$ is the smaller resistance, so it takes more: $12\ \text{W}$, against $6\ \text{W}$ for $X$. This is option (4).

Checks

$e - e * r / (a * b / (a + b) + r)$

$v * v / b$

Does it make sense?

Add the powers: $6 + 12 = 18\ \text{W}$. The battery gives out $EI = 12 \times 3 = 36\ \text{W}$. The rest, $36 - 18 = 18\ \text{W}$, is wasted as heat inside the battery: $I^2r = 3^2 \times 2 = 18\ \text{W}$. It all adds up.

The devices get less than the full $12\ \text{V}$, as they should, because the battery has resistance of its own.

Conclusion

$P_X = 6\ \text{W}, \quad P_Y = 12\ \text{W}$

The two devices together are $2\ \Omega$. With the battery's $2\ \Omega$ the total is $4\ \Omega$, so $I = 3\ \text{A}$.

The voltage across the devices is $12 - 3 \times 2 = 6\ \text{V}$. Then $P_X = \dfrac{36}{6} = 6\ \text{W}$ and $P_Y = \dfrac{36}{3} = 12\ \text{W}$. The answer is (4).

A battery with internal resistance does not give its full $E$ to the outside. Find the current first, then $V = E - Ir$.

Devices side by side have the same voltage. Use $P = \dfrac{V^2}{R}$ for each: the smaller resistance takes more power.

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