10 : Electrical circuits

Question 1

Kerchiefs second law

Answer 1

In any circuit, the sum of the electromotive forces is equal to the sum of the p.d.s. Around a closed loop
- because of the law of conservation of energy
- total energy transferred to the charges is always equal to total energy transferred from the charges

Question 2

A series circuit

Answer 2

• only one path for the current, a single loop from one terminal of the source of e.m.f. Back to the other terminal
• current is the same in every position from kirchoffs first law
• as there is only one loop from kirchoffs second law e.m.f. is shared between components, the sum of p.d.s across the component is equal to the e.m.f.

Question 3

Parallel circuits

Answer 3

• Provides more than one possible path for the charges, how much charge flows down each path depends on the resistance Kirchoffs first law tells us that the current into each junction must equal current out of that junction the greater resistance the less current
• each parallel branch can be thought of as separate circuit kirchoffs second law tells us that around each loop e.m.f. Must equal p.d. So in each branch there is same voltage

Question 4

Resistors in series

Answer 4

Each additional resistor increases the length of the path taken by charges
R = +R1+ R2
Because
V = V1+ V2
And V=IR
And I is constant

Question 5

Resisters in parallel

Answer 5

• Total resistance decreases an additional resistor provides another path for current effectively increasing cross sectional area and so decreases resistance

R=1/R1+1/R2
Because
I=I1 + I2
I / V =l1/ V + I2/V
And V is constant

Question 6

internal resistance

Answer 6

Whenever there is a current in a power source, work has to be done by the charges as they move through the power source. In a chemical cell this work is due to reactions between the chemicals. In a solar cell it is due to the resistance of the materials of the cell.
- as a result some energy is ‘lost’ when there is a current in the power source and not all the energy transferred to the charge available in the circuit. The p.d. Measured at the terminals of the power source.(terminal p.d) is less than the actual e.m.f.

Question 7

E.m.f and terminal p.d. Relationship

Answer 7

E.m.f. = terminal p.d. + lost volts
Changing the current affects the lost volts and terminal p.d. Increasing the current means that more charges travel through the cell each second and so more work is done by the charges, increasing the lost volts with lowers terminal p.d.

Question 8

Lost volts equations

Answer 8

= I x r
r is internal resistance

Question 9

Potential dividers

Answer 9

Use potential dividers to divide the p.d. To give any value you require up to the maximum supplied from the power source
- circuit connected across one of the resistors in parallel. The p.d. Supplied to this circuit can be varies to any value from zero to maximum supplied from the power source, depending on the resistances of R1 and R2
- from kirchoffs second law the p.d. Across each resistor must always add up to the p.d. From the power source

Question 10

Ratio of resistances

Answer 10

The p.d. Across each resistor in s potential dividers depends on thei resistances.. if they have the same resistance then the p.d. Is shared equallly between them
- if one has twice the resistance of the other then this one will receive two/thirds of the total p.d.
- V1/V2 = R1/R2
- Vout = (R2/R1+R2)xVin

Question 11

Potentiometer

Answer 11

A variable resistor with a sliding contact. Adjusting this contact varies the p.d. Between two of the terminals, giving a variable Voiut
- when contact is moved towards one end meaning resistance to Vout increases then Vout increases until at one end Vout = Vin. When the contact is moved towards over end Vout decreases until at B it is zero