Marking Guidance for Practice Assessment for Electricity  1 (a) 𝑇 = 4  ×  0.002 = 0.008

𝑓 =1𝑇

𝑓 =1

0.008 𝑓 = 125  𝐻𝑧  

(b) You will only see half a wave on the screen (c) 𝑉!"# =

𝑉!"#$2

𝑉!"# =122

𝑉!"# = 8.5  𝑉   2 (a) 𝑅!

𝑅!=𝑅!𝑅!

412 =

𝑅30

𝑅 = 10  𝛺   (b) There is no change to the voltmeter reading. 3 (a) (i) 2  𝛺 (ii) 10  𝑉 (b) 3.5 joules of energy are supplied to each coulomb of charge 4 (a) The capacitor stores 1000 µC of charge when the potential

difference across the capacitor is 1 volt. (b) 𝑄 = 𝐶𝑉

𝑄 = 1000  ×  10!!  ×  80 𝑄 = 8  ×  10!!  𝐶  

(c) 𝐸 =12𝐶𝑉

!

𝐸 =12  ×  1000  ×  10

!!  ×   80 ! 𝐸 = 3.2  𝐽  

(d) The flash rate will decrease because there is a smaller charging current. This means it takes longer for the capacitor to reach 80 V so the lamp is off for longer.

5 (a) The resistance of the material decreases (b) Electrons (c) The energy gap between the valance band and the conduction

band in a metal is less that the energy gap between the valance band and the conduction band in a semiconductor.