Electricity section 5: 5.7 Problem solving with series and parallel circuits - practice exam calculations - questions with worked out answers

Doc Brown's Physics exam study revision notes: There are various sections to work through, after 1 they can be read and studied in any order.

ANSWERS to 5.7 Practice exam calculation QUESTIONS on series and parallel circuits

Q1

From circuit diagram 38 deduce ...

(a) What is the total resistance of the circuit?

The resistors are in series so R_total = 5.0 + 8.5 = 13.5 Ω = 14 Ω (2 sf)

(b) Predict what current the ammeter should show.

From Ohm's Law equation: V = IR,  so  I = V/R = 12.0/13.5 = 0.89 A (2 sf)

(c) If a third resistance of 3.5 Ω is connected between the 5 and 8.5 ohm resistors, what will be the total resistance  and current flow?

The resistors are in series so R_total = 5.0 + 8.5 + 3.5 = 13.5 Ω = 17 Ω (2 sf)

From Ohm's Law equation: V = IR,  so  I = V/R = 12/17 = 0.71 A (2 sf)

(d) What third resistance would have to added to the circuit to reduce the current flow to around 0.2 A?

R_total = V/I  =  12/0.2 = 60 Ω

R₃ = R_total - R₁ - R₂ = 60 -5.0 - 8.5 = 46.5 Ω = 47 Ω (2 sf)

 

Q2  

Study the diagram of circuit 36 carefully and for each question part explain or justify your answer.

(a) Predict the reading on voltmeter V₂?

When components are wired in series you can add up the individual p.d. values to give the total.

Therefore: V_total = V₁ + V₂ = 20.0 = 12.0 + V₂, so V₂ = 20.0 - 12.0 = 8.0 V

(b) What current will flow through each resistor?

The ammeter in the main circuit reads 5.0 A.

Since everything is wired in series, the current at any point in the circuit is the same.

Therefore 5.0 A flows through each resistor.

(c) Calculate the resistance of each resistor.

From Ohm's Law equation: R = V/I

R₁ = V₁/5.0 = 12.0/5.0 = 2.4 Ω

R₂ = V₂/5.0 = 8.0.0/5.0 = 1.6 Ω

(d) What is the total resistance?

Since they are wired in series, you can add up the consecutive resistances.

R_total = R₁ + R₂ = 2.4 + 1.6 = 4.0 Ω

Note:

(i) The larger the resistance the greater the p.d. across it.

(ii) A numerical check on the R values. R_total = V_total/I_total = 20.0/5.0 = 4.0 Ω

 

Q3

Study the diagram of circuit 37 carefully and for each question part, explain or justify your answer.

(a) Predict the ammeter reading for A₂.

The total current running through the parallel section of the circuit must equal that running through the rest of the circuit.

So, A₃ = A₁ + A₂,   so   A₂ = A₃ - A₁ = 5.0 - 2.0 = 3.0 A

(b) Calculate the value of the parallel resistances R₁ and R₂.

Resistors wired in parallel have the same p.d. across them, so using Ohm's Law equation ...

R₁ = V₁/I₁ = 20.0/2.0 = 10.0 Ω

R₂ = V₂/I₂ = 20.0/3.0 = 6.7 Ω

(c) Calculate the total resistance of the resistor section of the circuit.

BUT not by adding them up!

You have to use the total current flowing and the potential difference of the battery.

R_total = V_total/I_total = 20.0/5.0 = 4.0 Ω

Note:

(i) The larger the resistance the lower the current running through it (c).

(ii) The total resistance is actually much less than any of the individual resistances in parallel.

If the resistors were wired in series, the total resistance would 16.7 Ω.

Again, the water pipe analogy helps here - think of the water being able to through two pipes of similar diameter.

(iii) The formula for adding up two resistances in parallel is:

1/R_total = 1/_R1 + 1/_R2

1/R_total = 1/10 + 1/6.7 = 0.10 + 0.15 = 0.25

Therefore R_total = 1 / 0.25 =  4.0 Ω

 

Q4 If a circuit contains two resistors of 5.0 Ω and two of 3.5 Ω,  what is the total resistance in the circuit if they are all wired in series?

R_total = sum of all the resistances in series.

R_total = 5.0 + 5.0 + 3.5 + 3.5 = 17.0 Ω

Q5 Three 1.5 V batteries are wired in series with two identical bulbs wired in series.

(a) What is the total p.d. across the bulbs?

1.5 x 3 = 4.5 V  (added up in series)

(b) What is the p.d. across each bulb?

When wired in series, the total p.d. is divided up between the resistances.

Since the bulb resistances are the same, each will have the same p.d.

The p.d. across each bulb will be 4.5/2 = 2.3 V (2sf)

 

Q6 Suppose a lamp, a buzzer and three 1.5 volt batteries were wired in series with an ammeter and switch closed (circuit 46 below).

The resistance of the lamp is 10.0 Ω and the resistance of the buzzer is 5.0 Ω.

(a) What is the total p.d. across the lamp and buzzer?

1.5 x 3 = 4.5 V

(b) What is the total resistance of the lamp and buzzer?

They are wired in series, so you can just add them up.

R_total = R_lamp + R_buzzer = 10 + 5 = 15 Ω

(c) What reading would you expect on the ammeter? and what current flows through each component?

They are both wired in series, so same current passes through everything.

From Ohm's Law: I = V / R = 4.5 / 15 = 0.30 A

(d) Calculate the p.d. across (i) the lamp, and (ii) the buzzer.

From Ohm's Law: V = I x R

(i) lamp p.d.: V = 0.3 x 10 = 3.0 V

(ii) buzzer p.d.: V = 0.3 x 5.0 = 1.5 V

(e) How could you quickly check if you had made an error in (d)?

Its a series circuit, so the total p.d. should be 4.5 = 3.0 + 1.5

 

(f) If the 10 Ω lamp was replaced by a 20 Ω lamp, what might you notice in the performance of he circuit?

(i) the lamp would glow dimmer because of the greater resistance reducing the current flow.

(ii) although its resistance is unchanged, the buzzer wouldn't sound as loud because of the overall greater resistance in the circuit reducing the current flow.

(g) Suppose we now rewire the10.0 Ω lamp and the 5.0 Ω buzzer in parallel (circuit 47 below).

The ammeter gave a reading of 1.35 A.

(i) Why is the current flow greater than when the lamp and buzzer were wired in series?

Resistors wired in parallel offer a smaller resistance to current flow.

Two parallel wires ('pipes') are available for the current to flow through.

(ii) What is the total resistance of the lamp and buzzer? Compare with the series circuit 46 above.

R = V / I = 4.5 / 1.35 = 3.33 Ω (3 sf)

Confirms your deduction in (i), the resistance is much less than 15 Ω, in fact it is less than any of the individual resistances, characteristic of a parallel circuit compared to the series circuits involving the same components.

(iii) If the current through the lamp is 0.45 A, what current flows through the buzzer?

For a parallel circuit: I_total = I_lamp + I_buzzer = 1.35 A

Therefore I_buzzer = 1.35 - 0.45 = 0.90 A

From the information you can also calculate this from Ohm's Law

I = V / R = 4.5 / 5.0 = 0.90 A

(iv) If a second lamp, wired in parallel with the first lamp, was added to the circuit, what changes might, or might not happen in the p.d. across the resistors and maximum current flow?

The p.d. across the resistors remains the same.

However, the current flow will increase because you now have a 3rd pathway for the current to flow through, and the total resistance is reduced.

(h) Explain the use of a component you add could to the circuit to vary simultaneously the brightness of the lamp and the volume of sound from the buzzer, and what is its symbol?

You can add a variable resistor to the circuit, wired in series with the parallel sections of the lamp and buzzer.

By increasing/decreasing the resistance you can decrease/increase the brightness of the bulb and loudness of the buzzer.

 

Q7 A 12 Ω resistor is wired in series with a 36 Ω and connected to a 24 V power supply.

(a) What is the potential difference across each resistor?

The p.d. across a resistor in a series circuit is proportional to its resistance.

The total resistance is 48 Ω

For the 12 Ω resistor: p.d. = 24 x 12 / 48 = 6 V

For the 36 Ω resistor: p.d. = 24 x 36 / 48 = 18 V

Maths check: V_tot = V_12Ω + V_36Ω = 24 V

(b) What single resistor could you use to replace the original pair?

For a series circuit: R_tot = 12 + 36 = 48 Ω, therefore a single 48 Ω resistor will do.

(c) What current flows through each resistor?

Since they are wired in series, they both experience the same current flow.

I = V / R = 24 / 48 = 0.50 A