Booklet 3 - Electricity (Worked Examples) Flashcards

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1
Q

Use the oscilloscope trace to calculate

(a) the frequency
(b) the peak voltage

A

(a) The distance between crests is 4 cm.
The time base is set at 5 ms cm–1
The period of the wave is
T = 4 x 5ms = 4 0.005s = 0.02 s
f = 1/T
f = 1/0.02
f = 50 Hz

(b) The height from bottom to top is 8 cm.
The height from middle to top is ½ × 8 = 4 cm
The volts/div was set at 2 V cm–1,
The peak voltage is:
Vpeak = 4 × 2
Vpeak = 8 V

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2
Q

A power pack is labelled with a voltage of 12 V r.m.s.
Calculate the peak voltage.

A

Vpeak = √2 x Vrms
Vpeak = 1.41 x 12
= 17.0 V

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3
Q

Example

Calculate the current flowing during a bolt of lightning which lasts for 10 ms and transfers 50 C of charge.

A

Q = 50 C

I = ?

t = 10 ms = 10 x 10-3 s

Q = It

50 = I x 10 x 10-3

I x 10 x 10-3 = 50

I = 50/10 x 10-3

I = 5000 A

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4
Q

Example

Calculate the current flowing when a 2 kΩ resistor is connected to a 6V battery.

A

V = 6 V

I = ?

R = 2 kΩ = 2 x 103 Ω

V = IR

6 = I x 2 x 103

I x 2 x 103 = 6

I = 6/2 x 103

I = 3 x 10-3 A

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5
Q

Example

Calculate the current drawn when a 2 kW kettle is plugged into the mains.

A
P = 2 kW = 2000 W
I = ?
V = 230 V (UK Mains Voltage - see previous LO)

P = IV
2000 = I x 230
I x 230 = 2000
I = 2000/230
I = 8.70 A (to 3 sig fig)

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6
Q

Example

Calculate the current flowing when 3500 W of power are supplied to a 100 Ω resistor.

A
P = 3500 W
I = ?
R = 100 Ω

P = I2R
3500 = I2 x 100
I2 x 100 = 3500
I2 = 3500/100
I2 = 35
I = √35
I = 5.92 A (to 3 sig fig)

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7
Q

Example

A 100 Ω resistor is attached to a 9 V battery. Calculate the power supplied.

A
P = ?
V = 9 V
R = 100 Ω
P = V<sup>2</sup>/R
P = 9<sup>2</sup>/100
P = 81/100
P = 0·81 W
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8
Q

Example

Calculate the total resistance of the circuit shown.

A
R<sub>T</sub> = ?
R<sub>1</sub> = 1·5 kΩ = 1500 Ω
R<sub>2</sub> = 100 Ω
R<sub>T</sub> = R<sub>1 </sub>+ R<sub>2 </sub>+ ....
R<sub>T</sub> = 1500 + 100
R<sub>T</sub> = 1600 Ω
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9
Q

Example

Calculate the total resistance of the circuit shown.

A
RT = ?
R1 = 5 Ω
R2 = 20 Ω
1/R<sub>T</sub> = 1/R<sub>1 </sub>+ 1/R<sub>2 </sub>+ ....
1/R<sub>T</sub> = 1/5 + 1/20
1/R<sub>T</sub> = 1/4
R<sub>T</sub> = 4/1 = 4 Ω
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10
Q

Example

Calculate the total resistance of the circuit shown.

A

25 & 25 in series

RT = ?
R1 = 25 Ω
R2 = 25 Ω
R<sub>T</sub> = R<sub>1 </sub>+ R<sub>2 </sub>+ ....
R<sub>T</sub> = 25 + 25
R<sub>T</sub> = 50 Ω

50 (above) in parallel with 100

RT = ?
R1 = 50 Ω
R2 = 100 Ω
1/R<sub>T</sub> = 1/R<sub>1 </sub>+ 1/R<sub>2 </sub>+ ....
1/R<sub>T</sub> = 1/50 + 1/100
1/R<sub>T</sub> = 3/100
R<sub>T</sub> = 100/3 = 33.3 Ω (to 3 sig fig)
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11
Q

A cell of emf 1.5 V is connected in series with a 28 Ω resistor. A voltmeter measures the
voltage across the cell as 1.4 V.
Calculate:
(a) the internal resistance of the cell
(b) the current if the cell terminals are short circuited
(c) the lost volts if the external resistance R is increased to 58 Ω.

A

(a) E = 1.5 V
R = 28 Ω
V = 1.4 V

In this case we do not know the current, I, but we can work it out:
V = IR
1.4 = I x 28
I = 0.05 A

E = V + Ir
1.5 = 1.4 + 0.05 x r
r = 2 Ω

(b) Short circuit: R = 0, V = 0; Assume E (1.5 V) and r (2 Ω) constant
E = V + Ir
1.5 = 0 + I x 2
I = 0.75 A

(c) Assume E (1.5 V) and r (2 Ω) constant
Combine E = V + Ir and V = IR to get E = I(R + r)
1.5 = I x (58 + 2)
I = 1.5/60 = 0.025 A
lost volts = Ir
= 0.025 × 2
= 0.05 V

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12
Q

Example

Calculate V1 in this circuit.

A
V<sub>1</sub> = ?
R<sub>1</sub> = 100 Ω (This is R<sub>1</sub> because it has been marked as V<sub>1</sub> in the question)
R<sub>2</sub> = 150 Ω
V<sub>s</sub> = 10 V
V<sub>1</sub> = R<sub>1</sub>/(R<sub>1</sub>+R<sub>2</sub>) x V<sub>s</sub>
V<sub>1</sub> = 100/(100+150) x 10
V<sub>1</sub> = 4 V

(NB The formula sheet uses V2 = R2/(R1+R2) x Vs but this can be adapted as above as long as the voltage and resistor in bold agree with each other i.e.
V1 = R1/(R1+R2) x Vs
or
V2 = R2/(R1+R2) x Vs

The choice of which resistor is labelled 1 or 2 doesn’t matter unless it has already been marked in the Q)

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13
Q

Example

Calculate V1 in this circuit.

A

V1 = ?
V2 = 3 V
R1 = 12 kΩ
R2 = 6 kΩ
​(Can leave R1 and R2 in kΩ since the formula is a ratio.)

V<sub>1</sub>/V<sub>2</sub> = R<sub>1</sub>/R<sub>2</sub>
V<sub>1</sub>/3 = 12/6
V<sub>1</sub> = 3x12/6
V<sub>1</sub> = 6 V
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14
Q

A circuit is attached to XY which has a resistance of 10 kΩ. Calculate the effective
resistance between XY and thus the voltage across XY.

A

The resistance between XY comes from the two identical 10 kΩ resistors in parallel.
The combined resistance is therefore 5 kΩ (You can check this using 1/RT = 1/R1 + 1/R2).
Use this resistance as R1 in the potential divider formula.
R1 = 5 kΩ
R2 = 20 kΩ
VS = 12 V
V1 = ?

V<sub>1</sub> = R<sub>1</sub>/(R<sub>1</sub> + R<sub>2</sub>) x V<sub>s</sub>
V<sub>1</sub> = 5/(5 + 20) x 12
V<sub>1</sub> = 2.4 V
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15
Q

A capacitor stores 4 x 10-4 C of charge when the potential difference across it is 100 V.

(a) Calculate the capacitance.
(b) Calculate the energy stored in the capacitor.

A
C = ?
Q = 4 x 10<sup>-4</sup> C
V = 100 V

(a) C = Q/V
= 4 x 10-4/100
= 4 x 10-6 F

(b) E = ½QV
E = ½ x 4 x 10-4 x 100
E = 0.02 J

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16
Q

A 40 µF capacitor is fully charged using a 50 V supply.

Calculate the energy stored in the capacitor.

A
E = ?
C = 40 µF = 40 x 10<sup>-6</sup> F
V = 50 V
E = ½ CV<sup>2</sup>
E = ½ x 40 x 10<sup>-6</sup> x 50<sup>2</sup>
E = 0.05 J
17
Q

A 50 µF capacitor stores 2 mC of charge when fully charged.

Calculate the energy stored in the capacitor.

A
E = ?
Q = 2 mC = 2 x 10<sup>-3</sup> C
C = 50 µF = 50 x 10<sup>-6</sup> F
E = ½ Q<sup>2</sup>/C
E = ½ x (2 x 10<sup>-3</sup>)<sup>2</sup> / 50 x 10<sup>-6</sup>
E = 0.04 J