2022 A/L Physics Past Paper · Question 30 (English Medium)

Answer: (3) $0.4\ \text{A}$

Given information

What we know:

The batteries are $12\ \text{V}$ and $8\ \text{V}$ with no internal resistance. In the figure both have their positive plates towards the two lower corners. So they push against each other, and the $12\ \text{V}$ one wins.

The resistors are $3\ \Omega$ and $4\ \Omega$ in the middle, $6\ \Omega$ along the bottom, and $4\ \Omega$ and $2\ \Omega$ along the top. The capacitor is $10\ \mu\text{F}$.

The circuit has reached steady state. The capacitor is fully charged.

What the question asks

We want: the current $I_3$ in the $3\ \Omega$ resistor.

What each letter means:

$I_1$ ($\text{A}$): the current along the top path through the $4\ \Omega$ and $2\ \Omega$ resistors.

$I_2$ ($\text{A}$): the current in the $6\ \Omega$ resistor.

$I_3$ ($\text{A}$): the current in the $3\ \Omega$ and $4\ \Omega$ resistors in the middle.

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

$R$ ($\Omega$): a resistance.

Relevant theory

The reason why the microammeter now indicates zero is that when the capacitor is charged to the maximum, the circuit gets disconnected.

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

The algebraic sum of the electromotive forces of a closed circuit is equal to the algebraic sum, of the products of the currents flowing along all branches of the circuit and their respective resistances

For example, a bucket with a closed bottom is under a tap. Water runs in while it fills. Once it is full no more water runs in, and the tap-pipe carries nothing.

The capacitor is like that bucket. In steady state it is full, so no current passes through it. The capacitor branch can be taken as a broken wire.

Then the $4\ \Omega$ and $2\ \Omega$ on top carry the same current $I_1$. The $3\ \Omega$ and $4\ \Omega$ in the middle carry the same current $I_3$. At a lower corner the current $I_3$ splits into $I_1$ and $I_2$.

$$I_3 = I_1 + I_2$$

In words: the current going into a corner equals the currents going out.

For a closed loop, the sum of the e.m.f.s equals the sum of the currents times their resistances.

$$\Sigma E = \Sigma I R$$

In words: the net e.m.f. in the loop is used up in the resistors.

Required equations

$I_3 = I_1 + I_2$

$0 = (4 + 2)\,I_1 - 6\,I_2$

$12 - 8 = (3 + 4)\,I_3 + 6\,I_2$

Applying it to this question

Volts equal ohms times amperes, so every term in the loop equations is in $\text{V}$.

$1$

$6\,I_1 = 6\,I_2 \Rightarrow I_1 = I_2$

$2$

$I_3 = I_1 + I_2 = 2\,I_2$

$4$

$12 - 8 = 7 \times 2\,I_2 + 6\,I_2 \Rightarrow 4 = 20\,I_2$

$0.2$

$I_2 = \dfrac{4}{20} = 0.2$

$0.4$

$I_3 = 2\,I_2 = 2 \times 0.2 = 0.4$

Reasoning and calculation

The current in the $3\ \Omega$ resistor is twice the current in the $6\ \Omega$ resistor, so $I_3 = 0.4\ \text{A}$.

That is option (3).

Checks

$v / (a + b + c)$

Does it make sense?

The net e.m.f. is $12 - 8 = 4\ \text{V}$. The total resistance is $3 + 4 + 3 = 10\ \Omega$. So the current is $\dfrac{4}{10} = 0.4\ \text{A}$, in agreement with the loop equations.

The current is far less than $\dfrac{12}{3} = 4\ \text{A}$ because the $8\ \text{V}$ battery works against the $12\ \text{V}$ one.

Conclusion

$I_3 = 0.4\ \text{A}$

In steady state no current passes through the capacitor. The loop equations give $I_2 = 0.2\ \text{A}$ and $I_3 = 2\,I_2 = 0.4\ \text{A}$.

The answer is (3).

In steady state a capacitor carries no current, so remove its branch and solve what is left.

Then use the junction rule and the loop rule one after the other.

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