PHYSICS_10_23_V1_TG_Ohm's law in complete circuits

Starter activity

(T) Teacher explanation. Derive the Ohm’s law to complete circuit

There are three ways to arrive at the equation relating EMF, terminal PD, current and internal resistance. It is worth discussing all three, to show their equivalence. The order you take will depend on the approach used previously with the class:

1. As charge goes around the circuit the sum of EMFs must equal the sum of voltage drops leading to:

ε = I R + I r

The terminal voltage is equal to I R so this can be rearranged to give:

V = ε – I r

and interpreted as terminal voltage = EMF – ‘lost volts’

2. Energy is conserved. Imagine a unit of charge, Q, moving around the circuit:

Q ε = Q I R + Q I r

This leads to the same equations as in (1) above.

3. Use Ohm’s law with E ‘driving’ current through the combined resistance (R + r):

I = ε / (R+r) ─ Ohm's law to complete circuit.

Activity. Compare EMF and internal resistance

Students put for each case <, > or =

(f) Formative assessment questions.

Ask learners to solve problems on calculation of electric circuits including the short circuit. Assess the consequences of short circuit and suggest how to protect electric circuits from short circuits.

1.         A 9.0 V battery has an internal resistance of 12.0 W.

(a)        What is the potential difference across its terminals when it is supplying a current of

50.0 mA?

(b)       What is the maximum current this battery could supply?

(c)        Draw a sketch graph to show how the terminal potential difference varies with the current supplied if the internal resistance remains constant. How could the internal resistance be obtained from the graph?

2.         A cell in a deaf aid supplies a current of 25.0 mA through a resistance of 400 W. When the wearer turns up the volume, the resistance is changed to 100 W and the current rises to 60 mA. What is the emf and internal resistance of the cell?

3.         Explain why the headlamps of a car go dim when the starter motor is used.

Answers and Worked Solutions

1. (a)    pd = E – I r = 9 – (50 x 10^-3 x 12) = 8.4 V

    (b)   Max current = E/r = 9 / 12 = 0.75 A

2.         E = I(R +r)

E = 25 x 10^-3 (400 + r)    and    E = 60 x 10^-3 (100 + r)

So

25 x 10^-3 (400 + r) = 60 x 10^-3 (100 + r) so r = 114.3 W

E = 10 + (25 x 10^-3 x 114.3) = 12.86 V

Self assessment.

Students assess yourself by using answer sheet and provide fair and helpful feedback

Students take a portion of the responsibility for monitoring their own progress