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Science Case study.

Electricity · Maharashtra Board STD 10 (MSBSHSE - English Medium). 5 questions.

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5 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
A boy noted the readings on his home’s electricity meter on Sunday at 8AM and again on Monday at 8AM (see figure below).
  1. What was the meter reading on Sunday?
  2. What was the meter reading on Monday?
  3. How many units of electricity have been used?
  4. In how much time these units have been used?
  5. If the rate is Rs. 5 per unit, what is the cost of electricity used during this time?
Answer
  1. 42919
  2. 42935
  3. 42935 - 42919 = 16 units
  4. 24 hours.
  5. Cost of 1 unit = Rs. 5
Cost of 16 units = 16 × 5 = Rs. 80
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Question 24 Marks
A student made an electric circuit shown here to measure the current through two lamps.
  1. Are the lamps in series or parallel?
  2. The student has made a mistake in this circuit. What is the mistake?
  3. Draw a circuit diagram to show the correct way to connect the circuit. Use the proper circuit symbols in your diagram.
Answer
  1. Lamps are in series.
  2. Student has connected ammeter in parallel with lamps. It should be connected in series.
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Question 34 Marks
Consider the circuits given below:
  1. In which circuit are the lamps dimmest?
  2. In which circuit or circuits are the lamps of equal brightness to the lamps in circuit?
  3. Which circuit gives out the maximum light?
Answer
  1. Circuit (ii).
  2. Circuit (iii).
  3. Circuit (iii).
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Question 44 Marks
The table below shows the current in three different electrical appliances when connected to the 240V mains supply:
Appliance Current
Kettle 8.5A
Lamp 0.4A
Toaster 4.8A
  1. Which appliance has the greatest electrical resistance? How does the data show this?
  2. The lamp is connected to the mains supply by using a thin, twin-cored cable consisting of live and neutral wires. State two reasons why this cable should not be used for connecting the kettle to the mains supply.
  3. Calculate the power rating of the kettle when it is operated from the 240V mains supply.
  4. A man takes the kettle abroad where the mains supply is 120V. What is the current in the kettle when it is operated from the 120V supply?
Answer
  1. Lamp; because least current is flowing through it.
  2. Large current drawn by the kettle; Earth connection needed.
  3. We know that
P = VI

V = 240V, I = 8.5A

P = 240 × 8.5 = 2040W = 2.04kW
  1. When connected to 240V supply, P = 2040W
R = $\frac{\text{V}_2}{\text{P}}=\frac{2402}{240}$

R = 28.23 ohm

Now, when V = 120V, R = 28.23 ohm

$\text{I}=\frac{\text{V}}{\text{R}}=\frac{120}{28.23}=4.25\text{A}$​​​​​​​
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Question 54 Marks
The figure given below shows an electric circuit in which current flows from a $6V$ battery through two resistors.
  1. Are the resistors connected in series with each other or in parallel?
  2. For each resistor, state the p.d. across it.
  3. The current flowing from the battery is shared between the two resistors. Which resistor will have bigger share of the current?
  4. Calculate the effective resistance of the two resistors.
  5. Calculate the current that flows from the battery.
Answer
$V = 6V, R_1 = 2 ohms, R_2 = 3 ohms$
  1. Resistor are connected in parallel.
  2. p.d. across each resistor is same and is equal to 6V.
  3. 2 ohms resistance have bigger share of current because of its lower resistance.
  4. Effective resistance = $R$
$\frac{1}{\text{R}}=\frac{1}{2}+\frac{1}{3}$

$\frac{1}{\text{R}}=\frac{5}{6}$

$R = 1.2\ ohms.$
  1. Current flowing through battery, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{1.2}=5\ \text{amps}.$
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