2022 A/L Physics Past Paper · Question 23 (English Medium)

Answer: (4) $2.0\ \text{A}$

Given information

What we know:

The battery has e.m.f. $E = 6.0\ \text{V}$.

The resistors are $R_1 = 3.0\ \Omega$ and $R_2 = 6.0\ \Omega$, in parallel. The switch $K$ is in the branch of $R_2$.

With $K$ open the ideal ammeter reads $I = 1.5\ \text{A}$.

The $3.0\ \Omega$ resistor alone would give $\dfrac{6.0}{3.0} = 2.0\ \text{A}$ from a perfect battery. The reading is lower, so the battery has an internal resistance.

What the question asks

We want: the ammeter reading $I^{\prime}$ when $K$ is closed.

What each letter means:

$E$ ($\text{V}$): the e.m.f. of the battery.

$r$ ($\Omega$): the internal resistance of the battery.

$V$ ($\text{V}$): the potential difference between the battery terminals.

$R_1$, $R_2$ ($\Omega$): the two resistors.

$I$, $I^{\prime}$ ($\text{A}$): the ammeter reading with $K$ open and with $K$ closed.

Relevant theory

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

V =E-Ir

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

The algebraic sum of the currents flowing across a junction of a circuit is zero

For example, think of a pump pushing water round a pipe. If the pump has some friction inside, the pressure at its outlet is a little less than what the pump makes. When a second pipe is opened more water flows, so more is lost inside the pump and the outlet pressure drops a little more.

A battery behaves the same way. Inside it is the internal resistance $r$.

$$V = E - Ir$$

In words: the terminal potential difference is the e.m.f. minus the potential drop inside the battery.

Both resistors are joined across the same two terminals, so each has the same potential difference $V$. Each carries $\dfrac{V}{R}$, and the ammeter carries their sum.

$$I = \dfrac{V}{R_1} + \dfrac{V}{R_2}$$

In words: the current from the battery is the sum of the currents in the two branches.

Required equations

$V = E - Ir$

$V = I R_1$

$I^{\prime} = V^{\prime}\left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)$

$I^{\prime} = \dfrac{E\,g}{1 + r\,g} \quad \text{where } g = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

Applying it to this question

Volts divided by ohms give amperes, so $\text{V}\ \Omega^{-1} = \text{A}$.

$4.5$

$V = I R_1 = 1.5 \times 3.0$

$1$

$r = \dfrac{E - V}{I} = \dfrac{6.0 - 4.5}{1.5}$

$0.5$

$g = \dfrac{1}{3.0} + \dfrac{1}{6.0} = 0.5$

$2$

$I^{\prime} = \dfrac{E\,g}{1 + r\,g} = \dfrac{6.0 \times 0.5}{1 + 1 \times 0.5} = \dfrac{3.0}{1.5}$

$4$

$V^{\prime} = E - I^{\prime} r = 6.0 - 2.0 \times 1.0$

Reasoning and calculation

With $K$ closed the potential difference across the resistors is $4.0\ \text{V}$. The $3.0\ \Omega$ resistor carries $\dfrac{4.0}{3.0} = 1.33\ \text{A}$ and the $6.0\ \Omega$ resistor carries $\dfrac{4.0}{6.0} = 0.67\ \text{A}$.

Together they carry $2.0\ \text{A}$, which is the ammeter reading. That is option (4).

Checks

$e * g / (1 + r * g)$

Does it make sense?

Closing $K$ gives the current a second path, so the reading must rise above $1.5\ \text{A}$.

A battery with no internal resistance would give $\dfrac{6.0}{2.0} = 3.0\ \text{A}$. Our $2.0\ \text{A}$ is lower, because some of the e.m.f. is lost inside the battery.

Conclusion

$I^{\prime} = 2.0\ \text{A}$

The internal resistance is $r = 1\ \Omega$. With $K$ closed the ammeter reads $I^{\prime} = 2.0\ \text{A}$.

The answer is (4).

When a battery gives a reading lower than $\dfrac{E}{R}$, it has an internal resistance. Find $r$ first from the open-switch reading.

Then use $V = E - Ir$ with the new total current.

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