Exam practice

Two cells in series have a total e.m.f. of 1.5 + 1.5 = 3 V, twice that of one cell. The p.d. across the bulb doubles, so by Ohm’s law ($I = V/R$) a larger current flows through it. Its power, $P = VI$, is then much larger, so it changes more electrical energy into light and heat every second, and glows more brightly.

Nichrome has a high resistance, so it turns a lot of electrical energy into heat; a high melting point (about 1400 °C), so it can glow red-hot without melting; and it doesn’t oxidize even when red-hot in air (900 °C and more). So it makes a heating element that heats quickly and lasts a long time.

In parallel, every appliance gets the full supply voltage (about 230 V) it is designed for; each can be switched on and off by its own switch without affecting the others; and if one stops working, the rest carry on. In series, the voltage would be shared, every appliance would have to be on at once, and one failure would stop everything.

Electric current is the rate of flow of charge: the amount of charge that passes a point in a circuit in one second. $I = Q/t$. Its SI unit is the ampere (A); 1 A = 1 C/s.

Electromotive force is the energy a source (a cell, battery or generator) gives to each coulomb of charge to drive it round the whole circuit. Its unit is the volt (1 V = 1 J/C).
It is not a force. A force is a push or a pull, measured in newtons; e.m.f. is an amount of energy per unit charge, measured in volts. It got the name “force” in the early days of electricity, before this was understood.

The e.m.f. of the cell is 1.5 V: the cell gives 1.5 joules of energy to each coulomb of charge that it drives round a circuit.

Ohm’s law: the current through a conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions stay the same.
If a current $I$ flows when the p.d. is $V$:
$$I \propto V \quad\Rightarrow\quad \frac{V}{I} = \text{constant}$$
This constant is the resistance $R$ of the conductor. So
$$\frac{V}{I} = R \quad\Rightarrow\quad V = IR$$

Electric resistance is the opposition a conductor offers to the flow of current through it. It is caused by the moving electrons colliding with the atoms of the conductor. $R = V/I$; its SI unit is the ohm (Ω).
1 Ω is the resistance of a conductor through which a current of 1 A flows when a p.d. of 1 V is applied across its ends: 1 Ω = 1 V/1 A.

The thin wire. Resistance increases as the cross-sectional area decreases: a thin wire gives the electrons a narrower path, so it opposes the current more.

The changing of electrical energy into heat when a current flows through a resistance is the heating effect of electric current. Examples: an electric heater and an electric iron (also a kettle, a rice cooker, an immersion rod).

Electric power is the electrical energy changed into other forms in one second: $P = E/t = VI$. Its unit is the watt (1 W = 1 J/s).
A rating of 5 W means the charger changes 5 joules of electrical energy every second when it works at full rating.

The bulbs were connected in series.
Reason: in series, the battery’s 3 V is shared among the bulbs. With three identical bulbs, each gets 3 V ÷ 3 = 1 V. With one taken out and the circuit joined up again, the same 3 V is shared between only two bulbs, so each gets 3 V ÷ 2 = 1.5 V. The total resistance is smaller (two bulbs instead of three), so a larger current flows, $I = V/R$, through each bulb. With more p.d. across it and more current through it, each bulb changes more electrical energy into light every second ($P = VI$), so it glows more brightly.
In parallel this wouldn’t happen: each bulb already has the full 3 V across it, so taking one out leaves the others glowing just as before.
(If the bulb had simply been unscrewed and the others had stayed lit without anything being rejoined, they could not have been in series, since a gap in a series circuit puts every bulb out. They would have to be in parallel, and then they would brighten only very slightly: with one bulb fewer, the cells give less current, so a little less of their energy is used up inside them and the p.d. across the bulbs rises a little.)

Join the two cells in series (the + of one to the − of the other), to get the largest p.d., 3 V. Then connect the two bulbs in parallel across them.
In parallel, each bulb gets the full 3 V, so each glows as brightly as one bulb alone on 3 V. (In series, each bulb would get only 1.5 V, and both would glow dimly.) The total light is greatest this way, though the cells run down sooner, because they give more current.

$I$ = 10 mA = 10/1000 A = 0.01 A, $Q$ = 0.2 C
$$I = \frac{Q}{t} \quad\Rightarrow\quad t = \frac{Q}{I} = \frac{0.2\ \text{C}}{0.01\ \text{A}} = 20\ \text{s}$$
20 s.

$I$ = 5 mA = 0.005 A, $t$ = 1 min = 60 s
$$Q = I \times t = 0.005\ \text{A} \times 60\ \text{s} = 0.3\ \text{C}$$
0.3 C.

$V$ = 3 V, $I$ = 500 mA = 0.5 A
Power: $P = VI$ = 3 V × 0.5 A = 1.5 W
Resistance: $R = V/I$ = 3 V ÷ 0.5 A = 6 Ω

Both are in parallel across the supply, so each has $V$ = 220 V.

Heater: $I = P/V$ = 1100 W ÷ 220 V = 5 A; $R = V/I$ = 220 V ÷ 5 A = 44 Ω

CFL: $I = P/V$ = 11 W ÷ 220 V = 0.05 A; $R = V/I$ = 220 V ÷ 0.05 A = 4400 Ω

The more powerful appliance has the smaller resistance: it lets more current through.

From the label, $P$ = 3000 W = 3 kW, and $V$ = 220 V.

Current: $I = P/V$ = 3000 W ÷ 220 V ≈ 13.6 A

Energy: $t$ = 4 min = 4/60 h

$E = P \times t$ = 3 kW × 4/60 h = 0.2 kWh = 0.2 units

Daily consumption, $E = P \times n \times t$ (with $P$ in kW and $t$ in hours):

• LED bulbs: 7/1000 × 10 × 6 = 0.42 kWh
• TV: 150/1000 × 1 × 8 = 1.2 kWh
• fans: 150/1000 × 4 × 10 = 6 kWh
• iron: 40 minutes a week = 40/60 h ÷ 7 ≈ 0.0952 h a day, so 750/1000 × 1 × 0.0952 ≈ 0.0714 kWh

Total in a day ≈ 0.42 + 1.2 + 6 + 0.0714 = 7.6914 kWh
In a month ≈ 7.6914 × 30 ≈ 230.7 units (kWh)
Bill ≈ 230.7 × Rs 10 ≈ Rs 2,307