Date | November 2019 | Marks available | 2 | Reference code | 19N.2.HL.TZ0.9 |
Level | Higher level | Paper | Paper 2 | Time zone | 0 - no time zone |
Command term | Calculate | Question number | 9 | Adapted from | N/A |
Question
X has a capacitance of 18 μF. X is charged so that the one plate has a charge of 48 μC. X is then connected to an uncharged capacitor Y and a resistor via an open switch S.
The capacitance of Y is 12 μF. S is now closed.
Calculate, in J, the energy stored in X with the switch S open.
Calculate the final charge on X and the final charge on Y.
Calculate the final total energy, in J, stored in X and Y.
Suggest why the answers to (a) and (b)(ii) are different.
Markscheme
E=12Q2C OR V=QC✔
E=«12(48×10-6)18×10-6=» 6.4×10-5 «J»✔
ALTERNATIVE 1
QX+QY=48 ✔
QX18=QY12 ✔
solving to get QX=29 «μC» QY=19 «μC» ✔
ALTERNATIVE 2
48=18 V+12 V⇒V=1.6 «V»✔
QX=«1.6×18=» 29 «QX=1.6×18=29 «μC» ✔
QY=«1.6×12=» 19 «μC» ✔
NOTE: Award [3] for bald correct answer
ALTERNATIVE 1
ET=12(29×10-6)218×10-6+12(19×10-6)212×10-6✔
=3.8×10-5«J»✔
ALTERNATIVE 2
ET=12×18×10-6×1.62+12×12×10-6×1.62 ✔
=3.8×10-5«J»✔
NOTE: Allow ECF from (b)(i)
Award [2] for bald correct answer
Award [1] max as ECF to a calculation using only one charge
charge moves/current flows «in the circuit» ✔
thermal losses «in the resistor and connecting wires» ✔
NOTE: Accept heat losses for MP2