Electricity & Magnetism

11 methods · 33 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

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Almost every numerical electricity question is a two or three step chain between $Q$ , $t$ , $I$ , $V$ , $R$ , $P$ and $E$ , and the current is the node every chain passes through. Before choosing a formula, ask which pair of given quantities delivers $I$ , get it in one line, then walk out to whatever is asked. Convert every time to seconds and every prefix to SI units at the moment you read them, never at the end.

Trigger: the question hands you a charge, a time in minutes, a power rating or a resistance, and asks for something else entirely.

Instances:
(i) charge and time given: $I = \frac{Q}{t}$ ;
(ii) power and voltage given: $I = \frac{P}{V}$ ;
(iii) power and resistance given: $I = \frac{P}{R}$ ;
(iv) voltage and total resistance given: $I = \frac{V}{R}$ ;
(v) then leave through $Q = I t$ , $V = I R$ , $P = V I$ or $E = P t$ .

Linked questions (3)

NSAA 2016 Section 1 Part B (ESAT Physics) · Q29

A heater is connected in series with a resistor and a

6.0 V

battery in the circuit shown.

The total resistance of the circuit is

15 Ω

. In

3.0

minutes,

180 J

of electrical energy is transferred into other forms in the heater.

How much charge flows through the heater in the

3.0

minutes and what is the voltage across the heater?

A circuit diagram. A 6.0 V battery sits at the top, connected by a wire down the right side to a resistor (drawn as a rectangle, labelled 'resistor'), which connects at the bottom to a heater (drawn as a row of three small rectangles, labelled 'heater'), which connects back up the left side to the battery: battery, resistor and heater all forming a single series loop.

A $1.2 C$ , $150 V$
B $1.2 C$ , $216 V$
C $7.5 C$ , $0.041 V$
D $7.5 C$ , $24 V$
E $72 C$ , $0.40 V$
F $72 C$ , $2.5 V$
G $450 C$ , $0.40 V$
H $450 C$ , $2.5 V$

Worked solution

Answer:F

The heater, resistor and battery are all in series, so the same current

I

flows through the heater as through the whole circuit. With total resistance

15 Ω

and battery voltage

6.0 V

I = \frac{V}{R _{total}} = \frac{6.0}{15} = 0.40 A .

3.0

minutes (

180 s

), the charge through the heater (the same as through every component, in a series circuit) is

Q = I t = 0.40 \times 180 = 72 C .

The power dissipated in the heater is

P = \frac{E}{t} = \frac{180}{180} = 1.0 W

, so the voltage across the heater is

V_{heater} = \frac{P}{I} = \frac{1.0}{0.40} = 2.5 V .

(Check: this leaves

6.0 - 2.5 = 3.5 V

across the resistor, a resistance of

3.5/0.40 = 8.75 Ω

; adding the heater's own

2.5/0.40 = 6.25 Ω

gives

15 Ω

, matching the total given.)

The answer is F.

NSAA 2017 Section 1 Part B (ESAT Physics) · Q22

A charge of

60 C

passes through an electrical component when a steady voltage of

10 V

is applied across it for

30

seconds. The resistance of the component remains constant throughout.

What is the resistance of the component?

A $0.050 Ω$
B $0.20 Ω$
C $0.50 Ω$
D $2.0 Ω$
E $5.0 Ω$
F $20 Ω$

Worked solution

Answer:E

The current is

I = \frac{Q}{t} = \frac{60}{30} = 2.0 A

.

The resistance is

R = \frac{V}{I} = \frac{10}{2.0} = 5.0 Ω

.

The answer is E.

NSAA 2017 Section 1 Part B (ESAT Physics) · Q30

A kettle is designed to work from a car's power socket. The kettle has a power rating of

150 W

when a constant voltage of

12.0 V

d.c. is applied across its element.

How much charge passes through the element of this kettle when the voltage of

12.0 V

is applied across it for

20

minutes?

A $96 C$
B $250 C$
C $15000 C$
D $36000 C$
E $900000 C$
F $2160000 C$

Worked solution

Answer:C

The current in the element is

I = \frac{P}{V} = \frac{150}{12.0} = 12.5 A

.

In

20

minutes (

1200 s

), the charge that flows is

Q = I t = 12.5 \times 1200 = 15000 C .

The answer is C.