1 · $Q=It$
Charge through a lamp
A current of $250\,\text{mA}$ flows for $2.0\,\text{minutes}$. Calculate the charge.
$$I = 0.250\,\text{A},\quad t = 120\,\text{s}$$ $$Q = It = 0.250 \times 120 = 30\,\text{C}$$2 · $E=QV$
Energy from charge and voltage
$30\,\text{C}$ passes through a $12\,\text{V}$ lamp. $E = QV = 30 \times 12 = 360\,\text{J}$.
3 · $V=IR$
Finding resistance
$V=6.0\,\text{V}$, $I=0.25\,\text{A}$. $R=V/I=6.0/0.25=24\,\Omega$.
4 · Series pair (2.19)
Two resistors on a $9.0\,\text{V}$ battery
$R_1=10\,\Omega$, $R_2=20\,\Omega$.
$$R_{\text{T}}=30\,\Omega,\quad I=9.0/30=0.30\,\text{A}$$ $$V_1=3.0\,\text{V},\quad V_2=6.0\,\text{V}$$5 · $P=IV$ and a fuse
Kettle on mains
$P=2.3\,\text{kW}$, $V=230\,\text{V}$.
$$I = P/V = 2300/230 = 10\,\text{A}$$Choose a $13\,\text{A}$ fuse. A $5\,\text{A}$ fuse would melt in normal use.
6 · $E=IVt$
Heater left on
$I=8.0\,\text{A}$, $V=230\,\text{V}$, $t=10\,\text{min}=600\,\text{s}$.
$$E = 8.0 \times 230 \times 600 = 1.104 \times 10^6\,\text{J}$$7 · Parallel current
Junction rule
Total current into a junction is $1.8\,\text{A}$. One branch takes $0.7\,\text{A}$. The other takes $1.1\,\text{A}$. Current is conserved — it is not used up.
8 · Explain (marked)
Why a metal case is earthed
If the live wire touches the case [1], a large current flows through the earth wire [1], the fuse melts [1], and the case is disconnected from live so the user does not get a shock [1].
“The earth wire takes the electricity away” with no fuse and no live-to-case story is 0 or 1.
Then sit the five quizzes and the three papers.