Question on variable capacitor

 

Electricity revision

 

Resistance calculations

 

1.       A motor has a resistance of 12 Ω. If the supply voltage is 6 V calculate the current flowing through it.

I = V/R = 6/12 = 0.5A

2.       A current of 0.55 A flows through a resistor of value 22 Ω. Calculate the voltage across the resistor.

V = IR = 0.55 x 22 = 12.1V

3.       See circuit diagram opposite. A bulb has a resistance of 9 Ω. The reading on the ammeter is   0.17 A. Calculate the reading on the voltmeter.

V = IR = 0.17 x 9 = 1.53V

4.       A motor has resistance of 210 Ω. A voltmeter placed in parallel with the motor reads 5 V. calculate the current flowing through the motor.

I= V/R = 5/210 = 0.024A

5.       A power supply is set to 9V. Calculate the current flowing through a 1.2 kΩ resistor.

1.2kΩ = 1200Ω Using I=V/R = 9/1200 = 0.0075A

6.       A torch bulb is connected to a 3V battery. It has a resistance of 8.5 Ω. Calculate the current flowing through it.

I = V/R = 3/8.5 = 0.35A

7.       A light bulb is connected to a 240V supply. A current of 0.24 A flows throug it. Calculate the resistance of the bulb.

R=V/I = 240/0.24 = 1000Ω

8.       An electric drill is connected to a 240V supply. A current of 4 A flows through it. Calculate its resistance.

R=V/I = 240/4 = 60Ω

 

9.       The reading on the ammeter in the circuit opposite is 1.69 A. The reading on the voltmeter is 10.9 V. Calculate the resistance of the bulb.

R=V/I = 10.9/1.69 = 6.5Ω

Ohm's Law

1 Find the current through a resistor of 20 Ω if the potential difference across it is 0.5 V.

I = V/R = 0.5/20 = 0.025A = 25mA

2 When a p.d. of 2 V is applied across a resistor the current is 0.1 A. What is the value of the resistor? R= V/I = 2/0.1 = 20 Ω

3 A p.d. of 100 V is applied across a resistor of 2 kΩ. What is the current?

I = V/R = 100/2000 = 0.05A

4 What p.d, will produce a current of 2 mA through a resistor of 4 kΩ?

V = I R = 0.002 x 4000 = 8V

5 Find the value of a resistor if a p.d. of 2 V produces a current of 0.5 μA.

R = V/I = 2/0.5 x10^-6 = 4000000 = 4MΩ

6 Find the current through a 1.5 MΩ resistor when a p.d. of 4.5 V is applied.

I = V/R = 4.5/1.5 x10⁶ = 3 x 10^-6A

7 A lamp operated by a 240 V supply takes a current of 0.625 A. Find the resistance of the lamp.

R = V/I = 240 /0.625 = 384 Ω

8 A coil of very fine copper wire is found to take a current of 0.75 A when a p.d. of 4.5 V is applied. If the wire has a resistance of 1.5 Ω per metre, what length of wire is in the coil?

R = V/I = 4.5/ 0.75 = 6 Ω

Length = 6 Ω/1.5 Ωm^-1 = 4m

Electron Beam worksheet

 

1        (i)      beam current or intensity is reduced (1)
(because) fewer electrons are emitted (per sec) from the filament (1)
[or no beam as no electrons emitted if voltage of A
reduced enough (1)
(only)]

(ii)     electrons travel faster [or more kinetic energy] (1)
(because the force of) attraction to the anode is greater (1)                                    4

[7]

          2   (a)           KE  = 1.6 x 10-19 x 25 x 103;

         

(b)     A calculation to include:

1. K.E. = ½ x m x v2;

2. 4 x 10-15 = 0.5 x 9.1 x 10-31 x v2;

3. v = 9.4 x 107 ms^-1;                                                                                           3

.        

3        current = charge flow/ second or I = Q/t;
            = 9.0 × 1015/s × 1.6 × 10 –19C;
            = 1.4 × 10–3A;                                                                                                 3

Electron at Right angles to an E Field

 

Electrons in an E Field

 

Charged particle in a field