Electricity Calculation and Practical Electricity Question

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1. Three bulbs are connected in a circuit as shown below. The electromotive force (e.m.f) of the battery is 12.0 V.



(a) State what is meant by the electromotive force (e.m.f) of the battery is 12.0 V.
(b) Calculate
      (i) the effective resistance, R_{eff, bulb},of the three bulbs
      (ii) the current passing through the battery
      (iii) the potential difference across PQ, V_PQ
      (iv) the amount of electrical energy converted by the bulb L₁ to heat and light energy when 4.0 * 10³ C of charge passing through the bulb L₁
(c) Bulb L₂ is now taken from its socket. State and explain what happens to the brightness of bulb L₁ and L₃.
      (i) brightness of bulb L₁,
      (ii) brightness of bulb L₃

(a) Comment: A very common mistake is to simply write the definition of e.m.f without the reference to 12V. By doing so, you wont be awarded with any marks.

Answer:
12.0 J of work is done by battery to drive a coulomb of charge (a unit charge) around a complete circuit.
(b)(i) Analysis:
L₂ and L₃ are first in series, then in parallel with L₁

R_{eff, bulb} = (1/100 + 1/200)^-1
                  = 66.7Ω
(ii) Analysis:
Currrent passing through the battery is same as current for the whole circuit. Therefore, we have to find the total effective resistance for the whole circuit first, then use I = V/R to find the current.
If you are not confident enough, you can redraw the circuit as shown below.



I = V/R
   = 12 / (40 + 66.7)
   = 0.1125 = 0.113 A
(iii) Analysis:
By looking at the redrawn circuit at (ii), there are three methods to do this question.
Method 1: use V = IR formula but R here is the R_{eff, bulb}
Method 2: find p.d across 40Ω first using V = IR, then V_PQ = emf - p.d across 40Ω since they are in series.
Method 3: use potential divider as V_PQ is same as V_out in potential divider circuit

Method 1: V_PQ = IR_{eff, bulb} = 0.1125*66.7 = 7.5V
Method 2: V_40Ω = IR = 0.1125 * 40 = 4.5V
                  V_PQ = 12 – 4.5 = 7.5V
Method 3: V_PQ = 66.7 / (40 + 66.7) * 12 = 7.5V
(iv) V = W / Q
   7.5 = W / (4.0 * 10³)
     W = 3.0 * 10⁴J
(c) Analysis:
Whenever the question asks about the brightness of bulb, you have to talk about power as brightness depends on power! From there, you then decide whether it becomes dimmer, brighter or remains the same.

Remember that bulb L₂ is taken out from its socket. The question doesnot mention it is replaced with a wire or something else. Thus, once it is taken out, it becomes open there.

(i) Before:
P = V² / R = 7.5² / 100 = 0.5625W
After:
I_new = V / R = 12 / (40 + 100) = 0.0857 A
P_new = I²R = 0.08572 * 100 = 0.735W

Since P_new > P, so L₁ becomes brighter.
(ii) L₃ does not light up as it is an open circuit.

2. Kai Kai connects a circuit as shown below.



Assuming that the ammeter and voltmeter are ideal,
(a) (i) Calculate the effective resistance of the resistors when only S₁ is closed.
      (ii) Calculate the ammeter and voltmeter reading when only S₁ is closed.
      (iii) If only S₁ is left closed for 2.5 minutes, calculate the amount of electric charge that flows through the 100Ω resistor.
(b) Now, both S₁ and S₂ are closed, state and explain the change
      (i) in the ammeter reading.
      (ii) in the voltmeter reading.

(a)(i) Analysis:
Since only S₁ is closed, we can redraw the circuit as shown below. In doing electricity calculation, you should always simplify the circuit diagram to help you solving the question.



From the circuit diagram above, we can clearly see that 80Ω and 20Ω are in series first, then in parallel with 100Ω.
So, ¹⁄_Reff = ¹⁄₁₀₀ + ¹⁄_{(80 + 20)}
            R_eff = 50Ω
(ii) Analysis:
For ammeter reading, it is pretty straightforward by using V = IR

I = V/R = 50/50 = 1.0A

Analysis:
For voltmeter reading, there are two methods to do this.
Method 1: Since both branches have a total resistance of 100Ω, the current would simply divide itself into 2, which means each branch has a current of 0.5A. Then, use V = IR to find the pd across 80Ω.
Method 2: If you observe the circuit carefully, pf across the branches is actually same as emf, 50V. So, we can use potential divider to find pd across 80Ω.

Method 1: I_80Ω = I / 2 = 0.5A
                  V = IR = 0.5 * 80 = 40V
Method 2: V = 80/(80+20) * 50 = 40V
(iii) Q = It = 0.5 * 2.5 * 60 = 75C
A common mistake is to forget converting the time from minutes to seconds.
(b)Analysis:
Now both S₁ and S₂ are closed.
Take note that when S₂ is closed, it actually creates a short circuit around 80Ω resistor. This means current will no longer pass through it then to 20Ω. Instead, current will pass through S₂ then to 20Ω. So, we can redraw the circuit as shown below.
(i) From the circuit, you should see that efective resistance decreases and hence current increases.
(ii) voltmeter reads the pd across a wire => 0V



Answer:
(i) By closing switch S₂, there is a short circuit across the 80Ω resistor. It becomes an easier path for current to flow through S₂ as compared with 80Ω resistor so effective resistance reduces and hence current increases
(ii) By closing switch S₂, there is no potential difference between two connection points (i.e V = 0V). Thus, voltmeter reading decreases to zero

3.

State the effect on the two lamps L₁ and L₂ in the Figure above when
(a) K₁ is open and K₂ is closed.
(b) K₁ is closed and K₂ is open.
(c) Both K₁ and K₂ are closed.

Analysis:
Remember the diode only allows current to flow in one direction, which is known as forward biased direction. In this case, both diode only allow current to flow up.
Also, remember that
In forward biased, diode is treated as a wire;
In reverse biased, diode is treated as open circuit
If you cannot remember this, check this out.

Analysis in general:
If you look at the circuit carefully, you should notice that current from positive terminal of battery would flow through either K₁ switch or L₁ branch as doide there is forward biased:
      (1) K₁ is open => current flows through L₁ branch;
      (2) K₁ is closed => current flows through K₁ as it creats a short circuit

Then now look at the another part of circuit, you should notice that current would not flow through L₂ branch at all as diode there is reversed baised (same as open circuit). This means whether current flows back to the negative terminal depends on whether K₂ is open or closed:
      (1) K₂ is open, current will not flow back to the negative terminal;
      (2) K₂ is closed, current will flow back to the negative terminal;

If you dont follow the analysis above, look at analysis for each part of the question, you should be able to understand it.

(a) When K₁ is open, the current starts from positive terminal of battery can only flow through L₁ branch as diode there is forward biased (same as wire). Diode at L₂ is reverse biased (same as open circuit), thus current cannot flow through that branch. Luckily, K₂ is closed, thus, current flows back to the negative terminal of battery. A complete circuit formed.

Answer:
L₁ lights up but L₁ does not
(b) When K₁ is closed, it creates a short circuit across L₁ branch and hence current from positive terminal flows through K₁ switch instead of L₁ branch. Diode at L₂ is reverse biased (same as open circuit), thus current cannot flow through that branch. Also, since K₂ is open, current cannot flow back to the negative terminal. It is an open circuit overall.

Answer:
Both L₁ and L₂ do not light up
(c) Same explanation as (b). But this time, K₂ is closed, thus current flows back to the negative terminal of battery. However, look at the circuit carefully, current doesnot pass through any bulbs. It only flows through in the wire!

Answer:
Both L₁ and L₂ do not light up

4. A circuit is set up as shown below with a light dependent resistor (LDR). Its resistance ranges from 1200 to 3600Ω.



(a) Under what physical conditions will LDR be at the lowest and highest resistance respectively?
(b) When there is no light falling on the LDR, calculate
      (i) the reading shown by the ammeter
      (ii) the reading indicated by voltmeter V₁.
      (iii) the reading on voltmeter V₂
(c) Now, when light shining on LDR, state and explain what happen to
      (i) V₁;
      (ii) V₂

5. An unqualified electrician connects an electric fan rated "240 V, 600 W" at home. The domestic wiring is shown in the Figure below. The mains supply is at 240 V and L, N and E denote the live, neutral and earth wires respectively. The fan has metal casing.



(a) What does it mean by "an electric fan rated "240 V, 600 W"".
(b) State two mistakes the unqualified electrician made in wiring the fan in the Figure.
(c) As a safety precaution, a fuse is to be placed in the circuit.
      (i) Determine the correct rating for the fuse,
      (ii) Explain where the fuse should be inserted in the Figure.
(d) The fan is switched on for 3 hours. Calculate the
      (i) amount of charge passing through the fan,
      (ii) total cost of electricity if the cost per kWh of electrical energy is $0.90.

6. Andrew has bought an old metal lamp in a junk shop. He opens the plug and tries to check the wiring, which is shown below.

Explain what happens to the fuse and Andrew if he touches the metal case of the old metal lamp after he connects it into mains supply.

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