2016PPP1Q40 – Voltage and Power of Bulb Y when X is switched off - 2013PPP1Q40 heaters

2016PPP1Q40

Solutions: Option C

[There is another similar question in 2013 Nov Pure Physics P1Q40. The answer is Option C too. Refer to the last section of this post]

Refer to the 2 videos below. The 1st one is faster if you know that the bigger the resistance, by proportion, the bigger the potential difference of the component as it will take a larger portion of the emf wrt to the other components in series. This method is especially useful for MCQ.

If you are still unsure, you may put in values to find the I, V and P across the components. This will be more tedious and time-consuming.

Another similar question is 2013 Nov Pure Physics P1 Q40.

2010SPP2Q12 Fuel gauge using variable resistor

Solutions:

As the fuel level drops, the float which stays on the fuel surface will descend. The rod which is attached to the float will turn clockwise about the pivot X. As the rod turns, the resistance on the variable resistor increases. This increases the resistance of the circuit. Hence the current flowing through the circuit will decreases, causing the needle to deflect more to the left, indicating towards E (empty). Thus the reading on the fuel gauge decreases.

Solutions:

(i) Both the fixed resistor and sensor are in series.

Total effective resistance Re = 5000 + 1000 = 6000 ohms

V = IR

12 = I x 6000

I = 0.0020 A

Hence potential across Y,

V = IR (where I is constant in a series circuit)

= 0.0020 x 1000

= 2.0 V

(ii) When the temperature increases and the resistance of sensor Y decreases, the total effective resistance of the circuit decreases. Current flowing through the circuit increases.

Since V = IR, where the R of the fixed resistor is a constant 5000 ohms, as current I increases, the potential difference across the 5000 ohms resistor will increase.

Using P = I^2R and P = V^2/R to explain which component uses the highest power or energy.

Instead of P = IV to explain, it might be easier to explain using P = I^2R when the components are connected in series where current I is constant. Hence power P is directly proportional to R. The bigger the R, the more power it uses.

Likewise, if the components are connected in parallel, it will be easier to use P = V^2/R, as the potential difference is constant for the components connected in parallel. Hence, power P is inversely proportional to R. The smaller the R, the more power it uses.

SP2021Q8

PP2017Q11