Electrical Power, Energy & Heating Effect

PHYCurrent, Drift & Resistance

This subtopic covers Joule heating and the power delivered by appliances: how to get current and resistance from a bulb's wattage rating, compute energy in kWh and its cost, and apply the maximum-power-transfer condition for a cell. The core idea is that electrical energy converts to heat at rate $P=VI=I^2R=V^2/R$ , and energy is power multiplied by time.

ISC examines unit handling (W, kWh, J, cal) and the $R=r$ condition almost every year.

Electric power (three forms)

P = VI = I^{2}R = \dfrac{V^{2}}{R}

P

power (W),

V

voltage (V),

I

current (A),

R

resistance (

\Omega

). From a rating

P,V

: current

I=P/V

and resistance

R=V^2/P

Joule heat (electrical energy)

H = I^{2} R t = V I t = \dfrac{V^{2}}{R} t

H

heat in joules,

t

time in seconds. Convert to calories with

1\,\text{cal} = 4.18\,\text{J}

, i.e. divide joules by 4.18.

Energy in kilowatt-hours

E_{kWh} = P_{kW} \times t_{h}

P_{kW}

power in kilowatts,

t_h

time in hours. One unit on the bill equals 1 kWh; cost =

E_{kWh} \times

rate per unit.

Maximum power transfer

R = r, \qquad P_{max} = \dfrac{E^{2}}{4r}

E

EMF (V),

r

internal resistance (

\Omega

). At

R=r

the current is

I = E/(2r)

and exactly half the EMF appears across the load.

Current from cells in series

I = \dfrac{n E}{R + n r}

n

identical cells in series, each EMF

E

and internal resistance

r

, with external resistance

R

. Series EMFs add and internal resistances add.

Appliance ratings (e.g. 100 W, 220 V) give the resistance at working temperature as $R=V^2/P$ ; this differs from cold resistance.
1 kWh $= 1000\,\text{W}\times3600\,\text{s} = 3.6\times10^{6}\,\text{J}$ ; use kWh for billing and joules for heat-quantity problems.
Heat in calories = (heat in joules) / 4.18; never multiply by 4.18.
For maximum power transfer the load equals the internal resistance, $R=r$ ; the efficiency at this point is only $50\%$ .
Power in a series chain is largest in the largest resistor ( $P=I^2R$ , common $I$ ); in parallel it is largest in the smallest resistor ( $P=V^2/R$ , common $V$ ).
Energy consumed depends on time: doubling running time doubles energy and cost but not power.
When current is given for a fixed time, always convert minutes/hours to seconds before using $H=I^2Rt$ in joules.

Where the marks go

Mixing time units — using minutes or hours in $H=I^2Rt$ instead of seconds gives a wrong joule value.
Multiplying joules by 4.18 to get calories instead of dividing by 4.18.
Using cold/rated resistance interchangeably; the resistance found from $R=V^2/P$ is the hot (operating) value.
Assuming maximum power transfer means maximum efficiency — efficiency at $R=r$ is only $50\%$ .

How the board asks it

Numericalthe three forms of power and $R=V^2/P$
An electric bulb is rated $100\,\text{W},\ 220\,\text{V}$ . Calculate the current through it and its resistance at working temperature.
Numericalenergy in kilowatt-hours and cost
A heater of $1500\,\text{W}$ is used for $4$ hours daily for $30$ days. Find the electrical energy consumed in $\text{kWh}$ and the cost at $\textsf{Rs.}\ 5$ per unit.
Numerical $H=I^2Rt$ with second and calorie conversions
A current of $2\,\text{A}$ flows through a $10\,\Omega$ resistor for $5$ minutes. Calculate the heat produced, in joules and in calories.
Give reasonsthe maximum power transfer condition $R=r$
A cell of e.m.f. $E$ and internal resistance $r$ is connected to a variable load $R$ . State the condition for maximum power transfer to the load and give a reason why the efficiency at this condition is only $50\%$ .
Give reasonspower distribution in series versus parallel
Two bulbs marked $40\,\text{W}$ and $100\,\text{W}$ (both $220\,\text{V}$ ) are connected in series across the mains. Account for the fact that the $40\,\text{W}$ bulb glows brighter.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.