1

TOPIC 7Electric circuits

2

• Charges will flow to lower potential energy

• To maintain a current, something must raise the charge to higher potential

• This is normally a battery or generator

• Charge flowing through a resistance loses energy

• The battery must supply energy

• Electromotive force (EMF) is work done per unit charge moved from negative to positive terminal

• U/Q has units of Volts (Joule/Coulomb)

• EMF is really a potential, not a force!

EMF

3

• Ideal sources of EMF simply raise charge in potential

• Real batteries (etc) have internal resistance

• Terminal voltage then depends on current flowing

Internal Resistance

R r

VT

I • Current through internal resistance r implies potential drop across it, = I r

• Terminal voltage VT = – I r

• Current in external circuit is

IR r

E

4

Example 1

A battery has an emf of 1.60 V and an internal resistance of 0.15Ω. It is connected to a load resistance of 3.0Ω,

(a)What is the terminal voltage of the battery?

(b)How much power is (i) delivered to the load resistor; (ii) dissipated inside the battery; (iii) produced in total by the battery?

5

Resistors in series have the same current flowing through each resistor.

Voltage across resistor n is given by Vn = I Rn.

Total voltage is sum of individual voltages.

Overall resistance is total voltage divided by current

Total resistance is more than the largest individual value.

Resistors in Series

R1

I

R2

Tot 1 2 1 21 2

V V V IR IRR R R

I I I

6

Resistors in parallel have the same voltage across them.

Current through resistor n is given by In = V / Rn.

Total current is sum of individual currents.

Overall resistance is voltage divided by total current

Total resistance is less than the smallest individual value.

Resistors in Parallel

Tot 1 2 1 2

1 2

1 1 1V V

I I I R R

R V V V R R

R1 R2

7

A circuit composed only of resistors and a fixed emf will have a constant current flowing.

Where a capacitor is present, current flow will charge or discharge the capacitor, so the potential difference across its terminals will change.

As a result, the current through the rest of the circuit is likely to vary in time.

We will consider discharging and charging capacitors.

RC Circuits

8

Switch, initially closed, opened at t = 0

Current = rate of discharge

Potential difference

So

Exponential decay of the charge.

Discharging a Capacitord

d

QI

t

QV IR

C

d

d

Q QR

C t

00

1 dd

Qt

Q

Qt

RC Q

0

lnQ t

Q RC

0

tRCQ Q e

C R + +

– –

9

Time constant = RC. Charge dies to 1/e when t = .

T½ = ln(2)

Discharging a Capacitor (2)0

tRCQ Q e

t

Q

T ½

Q 0

0.5 Q 0

0.37 Q 0

Since V = Q/C, voltage decays in the same way

Since

Same time dependence again.

tRCV e

E

d

d

QI

t

00

t t tRC RC RC

QI e e I e

RC R

E

10

Switch closed at t = 0

Current = rate of charging

So

Hence

Exponential rise of the charge towards final value Q0.

Charging a Capacitord

d

QI

t

QIR

C E

0

0

lnQ Q t

Q RC

0 1t

RCQ Q e

C

R

+ –

I

0d

d

Q QQ Q C QR

t C C C

EE

11

Charging a Capacitor (2) 0 1

tRCQ Q e

t

Q

Q 0

0.63 Q 0

0 1 1t t

RC RCQQ

V e eC C

E

00

d

d

t t tRC RC RC

QQI e e I e

t RC R

E

12

Example 2

Initially, switch A is closed and B is open to charge C1. At time t = 0, switch A is opened and B is closed. How does the voltage across C2 vary with time?

C1

R C2

A B

13

We have looked at resistors (and capacitors) in series and in parallel.

Some circuits are more complicated than that!

Can use Kirchhoff’s junction and loop rules.

(Simple ideas, but usually complicated algebra! Take care with signs!)

Junction rule

Algebraic sum of all currentsentering & leaving a junctionis zero. I1 + I2 – I3 – I4 = 0

Just conservation of charge!

Kirchhoff’s Rules

I1 I3

I4 I2

14

Loop rule

The algebraic sum of the changes in potential around a closed loop is zero.

When any charge passes in a closed loop around a circuit, it returns to the same potential

Just conservation of energy!

Going from negative to positive terminal of a battery is a positive change in potential.

Going through a resistor in the same direction as the (assumed) current is a negative change in potential.

Going through a resistor in the opposite direction to the (assumed) current is a positive change in potential.

15

Example 3

(a) Apply Kirchhoff’s rules to the circuit above, to determine the current I flowing through resistor R1.

(b)Show that the same result can be obtained by calculating the effective resistance of the series-parallel combination of resistors.

A more realistic application of Kirchhoff’s rules is available on the web page.