1. (a)Ohms law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Using a voltmeter and ammeter to measure the voltage drop and the current flowing through each of the resistors the following results were obtained. From these values the value of each resistor was calculated using Ohm’s law V=IR. R1 | 43.75 | | V1 | | 3.5 | I1 | | 0.08 | | | R2 | 43.75 | | V2 | | 3.5 | I2 | | 0.08 | | | | | | | | | | | | | | R3 | 87.5 | | Vs | | 7 | I3 | | 0.08 | | | | | | | | | | | | | |

Were R3= total resistance in a circuit consisting of R1 and R2 in series. The value for each of the individual resistors can be obtained using ohms law I=V/R

∴R= V/I

∴R1=3.5/0.08= 43.75

∴R2=3.5/0.08= 43.75

R3 (total resistance) can also be calculated using Ohms law, where I is the current flowing through the total circuit and V is the voltage drop across both resistors.

∴R=V/I

∴R=7/0.08

∴R=87.5

This correlates exactly with the calculated value for total resistance in the results. It can be seen therefore that total resistance of a circuit with resistors in series does follow the law RT=R1+R2+…..Rn

(b) Resistors are said to be in parallel when they are connected to each other at both ends.

Using a voltmeter and ammeter to measure the voltage drop and the current flowing through each of the resistors connected in parallel the following results were obtained.

R1 | 35 | | V1 | | 7 | I1 | | 0.2 | | | R2 | 35 | | V2 | | 7 | I2 | | 0.2 | | | | | | | | | | | | | | R3 | 17.5 | | Vs | | 7 | I3 | | 0.4 | | | | | | | | | | | | | |

R1 and R2 were calculated using ohms law.

R =V/I

∴R1=7/0.2 = 35Ω

∴R2=7/0.2 = 35Ω

From the voltage measured across, and current flowing through, the circuit the total resistance (RT) can also be calculated.

RT=7/0.4

= 17.5 Ω

2.

(a) For 20Ω the circuit would be as follows:

(b)For 5Ω the circuit would be as follows:

3.

Using Kirchhoff’s law deduce the p.d across R and hence deduce the value of R.

Kirchhoff’s law states that: the sum of the currents flowing into a circuit junction must be equal to the sum of the currents flowing out of the junctions. From the diagram it can be seen that the total current flowing through the circuit is 0.1 A

From Ohms law I=V/RT

∴ 0.1 =10/(20+R)

∴20+R =10/0.1= 100

∴R = 100-20 = 80 Ω

4.

(a) The combined resistance of the two resistors in parallel.

1/RT =1/4.3 + 1/4.3

∴ 1/RT = 0.2326+0.2326

∴ RT = 1/0.5652

∴ RT = 1.77 Ω

(b) The total resistance of the circuit.

RT = 1.77 Ω +1.1 Ω

= 2.87 Ω

(c) The current in the circuit.

IT = V/RT

∴ IT= 6/2.87 Ω

∴ IT=2.09 amps

(d) Calculate VX and VY

VX= R(vx) /( R(vx)+R(vy)) X V in

∴ VX= 1.77/(1.77+1.1) X 6

∴ VX= (1.77/2.87) X 6

∴ VX= 0.617 X 6

∴ VX= 3.7 v

VY= 6v - VX

∴ VY= 6- 3.7

∴ VY= 2.3v

(e) Calculate current in each of the 4.3 Ω resistors.

I=Vx/R

∴I= 3.7/4.3

∴I= 0.860 amps in each of the 4.3 resistors.

5.

If two or more resistors are connected in series they can form a potential divider as follows:

6.

Calculate the potential difference across R1 when the resistance of T falls to 120 Ω

VT= supply voltage X RT/ (R1 +RT)

∴ VT= 10v x 120/(100+120)

∴ VT= 10v x 0.545

∴ VT= 5.45v

7.

A parallel plate capacitor has plates that you can slide closer or further apart. A high resistance voltmeter is connected across the plates and reads 20V when they are first disconnected from the power supply.

Assuming no charge leaks away, what would the voltmeter read when the distance between the plates is ….

(a) Halved

C = εr ε0 A d

Where C = Capacitance εr= Relative Permittivity of the plate material. ε0 = Electric Constant (Dielectric = permittivity of a vacuum = 8.854 x 10-12). A = Area of overlap of the two plates, (m2). d = Distance between the