Kirchhoff part 2

Starter The figure below shows a network of identical resistors.

2.0kΩ

2.0kΩ

2.0kΩ 2.0kΩ

A

B

Calculate the total resistance between points A and B.

resistance = ..................... kΩ

[Total 3 marks]

Starter The figure below shows a network of identical resistors.

2.0kΩ

2.0kΩ

2.0kΩ 2.0kΩ

A

B

Calculate the total resistance between points A and B.

resistance = ..................... kΩ

[Total 3 marks]

1.5Ω

Learning objectives

• State Kirchhoff’s second law• Apply Kirchhoff’s second law to circuits• Solve circuit problems involving series and

parallel circuits with one or more sources of e.m.f.

Kirchhoff’s Second Law

I am back!!This time it is slightly more complicated

What’s it all about?

Kirchhoff’s second law

• The sum of the e.m.f.s is equal to the sum of the p.d.s in a closed loop. This is an example of conservation of energy.

What it essentially says is that all the energy put into a circuit from the battery has to go somewhere. And this balance must be exact. You can't have even a small amount of energy appearing from nowhere or disappearing without trace.

V

R

r

Є V

ITerminal p.d. measured here

Lost volts across the internal resistance

e.m.f. = terminal p.d. + lost volts Є = V + v

We know that current through both resistors is ISo applying V= IR to each resistor

Є = V +Ir Є = IR + Ir Є = I(R+r)

6.0V

1Ω

7Ω4Ω

I

A battery of e.m.f. 6.0V and internal resistance 1Ω is connected to two resistors of 4Ω and 7Ω in series: calculatea. The total resistance in the external circuitb. The current supplied by the batteryc. The terminal p.d. of the battery

12V14V

0.04

0Ω

0.05

0ΩA 12V car battery is recharged by passing a current through it in the reverse direction using a 14V charger. Calculate the charging current

Note the 12 V e.m.f. opposes the 14V e.m.f. so: the sum of the e.m.f.s = 14 + - 12 = 2V

There are 2 internal resistors, and both ‘waste’ electrical energy

Є = I( R +r)