2018 A/L Physics Past Paper · Question 25 (English Medium)

Answer: (5) 180 mW

Given information

The source is $E = 10$ V. Its internal resistance is negligible.

The current is $I = 20$ mA, which is $0.02$ A.

A $50\ \Omega$ resistor is in series with $Z$. We do not know what is inside $Z$.

What the question asks

We want: the power dissipated in $Z$.

We do not need the resistance of $Z$. The voltage across it and the current through it are enough.

Relevant theory

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

This means that the electric charges flow without getting collected at any point or a junction of a circuit.

Required equations

$V_R = IR$

$E = V_Z + V_R$

$P_Z = V_Z I$

Applying it to this question

$1$

$V_R = 0.02 \times 50$

$9$

$V_Z = 10 - 1$

Reasoning and calculation

$P_Z = V_Z I = 9 \times 0.02$

$0.18$

$P_Z = 0.18$

$180$

$P_Z = 0.18 \times 1000$

$Z$ has $9$ V across it and $20$ mA through it. So it dissipates $180$ mW. That matches option (5).

Checks

$(E - I * R) * I * 1000$

$0.2$

$P = EI = 10 \times 0.02$

$0.02$

$P_R = I^2 R = 0.02^2 \times 50$

$0.18$

$P_Z = 0.2 - 0.02$

Volts times amperes gives watts.

$$0.18\ \text{W} = 180\ \text{mW}$$

The options are in mW too.

The source gives $200$ mW in all. The resistor takes $20$ mW.

So $Z$ gets $180$ mW. That makes sense.

Conclusion

$P_Z = 9 \times 0.02 = 0.18\ \text{W}$

$V_Z = 9$ V and $I = 20$ mA.

$$P_Z = 9 \times 0.02 = 0.18\ \text{W} = 180\ \text{mW}$$

The answer is option (5).

For a box you cannot see inside, do not look for its resistance.

Find the voltage across it and the current through it. Then use $P = VI$.

Exam tip