4TG_Electrical Power. Joule’s law. Current supply efficiency

OCR (A) specifications:

Start

Question: What is power dissipated in a circuit?

(G) Group work.

Divide learners into groups and ask each group to derive the formula for calculation of total and lost power and efficiency in an electric circuit. Then, ask learners to present their works at the blackboard.

(T) Teacher explanation.

• Effect of internal resistance on Power from a battery

Consider a simple circuit in which a battery of emf  and internal resistance  drives a current  through an external resistor of resistance . The external resistor is usually referred to as the load resistor. It could stand for either an electric light, an electric heating element, or, maybe, an electric motor. The basic purpose of the circuit is to transfer energy from the battery to the load, where it actually does something useful for us (e.g., lighting a light bulb, or lifting a weight). Let us see to what extent the internal resistance of the battery interferes with this process.

The equivalent resistance of the circuit is  r + R (since the load resistance is in series with the internal resistance), so the current flowing in the circuit is given by 

The power output of the emf is simply 

The power dissipated as heat by the internal resistance of the battery is 

Likewise, the power transferred to the load is 

Note that total power:

• Electric Power Efficiency

Power efficiency is defined as the ratio of the output power divided by the input power:

η = 100% ⋅ P_out / P_in

η is the efficiency in percentage (%).

P_in is the input power consumption in watts (W).

P_out is the output power or actual work in watts (W).

(P) Pair work.

Teacher pair up weak learners with more able students and they  may choose:

ü  Intermediate level

ü  Higher level

ü  Extension level

problems.

Marking scheme

Worksheet

1 a          W = VQ [1];    W = 1.5 × 1 = 1.5 J [1]

b                     W = 1.5 × 600 = 900 J [1]

2          A p.d. of 6.0 V across the lamp means that 6.0 J of electrical energy is transferred into heat and light per coulomb of charge flowing through the lamp. [1]

3          V = ^W_Q [1];  V = ₄¹⁵_.2 = 3.57 V ≈ 3.6 V [1]

5          1 kW h is the energy transferred by a 1 kW device operated for 1 hour. P = 1 kW = 1000 W, t = 1 hour = 3600 s [1]

1 kW h = Pt = 1000 × 3600 = 3.6 × 10⁶ J (3.6 MJ) [1]

6 a          Energy = 0.9 kW × 2.0 h [1];      energy = 1.8 kW h [1];

energy = 1.8 × 3.6 × 10⁶ ≈ 6.5 MJ [1]

b              Cost = 1.8 × 7.5p [1];      cost = 13.5p [1]

7          P = I ²R [1]

       I =      ^P =  ^0.25 [1]

R                                      100

I = 5.0 × 10^–2 A (50 mA) [1]

b                     0.36 = 0.8 × t (t = time in hours) [1]

t = ^0.36 = 0.45 h (t = 27 minutes) [1] 0.8

12      a   In a series circuit the current through the wires is the same. [1]

P                          = I²R = I² ( ^ρl ) ∝ ρ                                                           [1]

                                     A

The power dissipated is directly proportional to the resistivity ρ of the material.

Therefore, the nickel wire gets hotter. [1]

b                      The p.d. across the wires is the same when they are connected in parallel. [1]

The power dissipated is inversely proportional to the resistivity of the material.

Therefore, the iron wire will be hotter. [1]

(I)(E) Individual work.

Learners individually solve problems.

Marking scheme

End-of-chapter test

1          Difference: Charges are losing energy for p.d. and gaining energy for e.m.f. (for p.d., energy is transformed from electrical to heat, whereas for e.m.f., energy is transformed into electrical from other forms like chemical, etc.). [1]

Similarity: Both are measured in volts (both are defined as: ‘energy transferred per unit charge’). [1]

2      a     W = VQ [1];    W = 230 × 31 = 7.13 × 10³ J ≈ 7.1 × 10³ J [1]

b        P = ^W [1];  P = ^{7.13 × 103} ≈ 120 W [1]

t                                      60

3      a     P = I²R [1];   P = 0.25² × 100 [1];     P = 6.25 W ≈ 6.3 W [1]

b        P = VI [1]; V = ^P = ^2.0 = 8.0 V [1] I 0.25

4          a 1kWh is the energy transferred by a 1kW device operating for a period of 1hour. [1]

b     i      Energy = 0.085 × 5.0 [1]; energy = 0.425 kW h ≈ 0.43 kW h [1]

ii        Cost = 0.425 × 7.1 [1]; cost = 3.0p [1]

5          a P = ^V2 [1] R

b     i      R = ^V2 = ¹²² [1]; R = 24 Ω [1]

P         6.0

ii        For a given resistance, the power is directly proportional to the square of the p.d. ( P = ^{V 2} ∝ V ² ). [1] Hence the power dissipated by the resistor will

R

decrease by a factor of 4 if the p.d. is halved. [1]

_{ρ =} ^RA ₌ ^6.4 × ^7.0 × ^10–8 _{[1];  ρ = 8.96 × 10}–8 _{Ω m ≈ 9.0 × 10}–8 _{Ω m [1]}

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