Figure 13 shows an LED torch - AQA - GCSE Physics - Question 9 - 2020 - Paper 1

The equation linking charge flow, current, and time is:

$Q = I imes t$

Given that the current (I) is 50 mA, we first convert this to amperes:

$I = 50 ext{ mA} = 0.050 ext{ A}$

Now, using the formula:

$Q = I imes t = 0.050 ext{ A} imes 14400 ext{ s} = 720 ext{ C}$

Thus, the total charge flow through the cells is 720 C.

The torch did not work because when the cells were inserted the wrong way around, the current was unable to flow through the diode (LED). In a diode, current can only flow in one direction; if it is reversed, the diode blocks the current due to its high resistance.

The equation linking efficiency, total power input, and useful power output is:

$Efficiency = \frac{Useful \ power \ output}{Total \ power \ input}$

Given the total power input is 0.24 W and the efficiency is 0.75, we can calculate the useful power output:

$Useful \ power \ output = Efficiency \times Total \ power \ input$

Substituting the values:

$Useful \ power \ output = 0.75 \times 0.24 = 0.18 \text{ W}$

Thus, the useful power output of the LED is 0.18 W.