Class 9 Current Electricity exercise question answer science and technlogoy
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1. The accompanying figure shows some electrical appliances connected in a circuit in a house. Answer the following questions.
A. By
which method are the appliances connected?
Answer: Parallel connections.
B. What must be the potential difference across individual appliances?
Answer: The same as the supply voltage.
C. Will the current passing through each appliance be the same? Justify your answer.
Different in general . We have I = V/R .
Even if the voltage (V). is the same, the resistance (R) can be different.
Hence, the current (I) through each appliance may not be the same.
D. Why are the domestic appliances connected in this way?
Answer:
Even if one of the appliances is out of order, other appliances can be used.
E. If the T.V. stops working, will the other appliances also stop working? Explain your answer.
Answer:
Even if the TV stop working as they are
connected in parallel across the supply.
2.The following figure shows the symbols for components used in the accompanying electrical circuit. Place them at proper places and complete the circuit
Answer:
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Which law can you prove with the help of the above circuit?
Answer:
Ohm’s law : The electric current flowing in a metallic conductor is directly proportional to the potential difference across its terminals, provided the physical conditions of the conductor such as it’s length, area of cross section, temperature and material remain constant.
3. Umesh has two bulbs having resistances of 15 W and 30 W. He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.
A. Which method should he use to connect the bulbs?
Answer: parallel combination
B. What are the characteristics of this way of connecting the bulbs depending on the answer of question A above?
Answer: characteristics of a parallel combination of resistors:
- The voltage (potential difference) across each of the resistors in the same.
- The total current is equal to the sum of the current trough the individual resistors.
- The reciprocal of the effective resistance of the combination is equal to the sum of the reciprocals of the individual resistances.
- The effective resistance of the combination is less than the any individual resistances.
- The current through each resistor is inversely proportional to the resistance of the resistor.
- This combination can be used to decrease the resistance in a circuit.
C. What will be the effective resistance in the above circuit?
Answer:
4. The following table shows current in Amperes and potential difference in Volts.
|
V (Volts) |
I (Amp) |
|
4 |
9 |
|
5 |
11.25 |
|
6 |
13.25 |
a. Find the average resistance.
Answer:
R1 = 4V/9A = about 0.44 Ohm
R2 = 5V / 11.25A = about 0.44 Ohm,
R3 = 6V
/ 13.5 = about 0.44 Ohm
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b. What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
Ans: A
straight line passing through the origin ( 0 , 0 )
c. Which law will the graph prove? Explain the law.
Ans :
Ohm’s law
5. Match the pairs
‘A’
Group ‘B’ Group
1. Free
electrons a. V/ R
2. Current b. Increases the resistance in the circuit
3.
Resistivity c. Weakly
attached
4. Resistances in d. VA/LI Series
Answer:
- Free electrons – Weakly attached
- Current – V/ R
- Resistivity – VA/LI
- Resistances in Series – Increases the resistance in the circuit.
6. The resistance of a conductor of length x is r. If its area of cross section is a, what is its resistivity? What is its unit?
Answer:
In the usual notation
R = p × L / A
Resistivity p = RA / L
Here, p = ra/x
Unit:
OHm
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7. Resistances R1 , R2 , R3 and R4 are connected as shown in the figure. S1 and S2 are two keys. Discuss the current flowing in the circuit in the following cases.
a. Both S 1 and S 2 are closed.
b. Both S 1 and S 2 are open.
c. S 1
is closed but S 2 is open.
Answer:
See the
following figure:
R4 is shunted by S2. Hence the resistance of this combination will be practically zero.
R1 and
R2 in parallel:
2) see
the following figure:
3) See
the following figure:
8. Three resistances x1 , x2 and x3 are connected in a circuit in different ways. x is the effective resistance. The properties observed for these different ways of connecting x1 , x2 and x3 are given below. Write the way in which they are connected in each case. (I-current, V-potential difference, x-effective resistance)
a. Current I flows through x1 , x2 and x3
b. x is larger than x1 , x2 and x3
c. x is smaller than x1 , x2 and x3
d. The potential difference across x1 , x2 and x 3 is the same
e. x = x 1 + x 2 + x 3
f. x = 1 /1/ x 1 + 1/ x 2 + 1/ x3
Answer :
Series
combination
Series
combination
Parallel
combination
Parallel
combination
Series
combination
Parallel
combination
9. Solve the following problems.
A. The resistance of a 1m long nichrome wire is 6 W. If we reduce the length of the wire to 70 cm. what will its resistance be? (Answer : 4.2 W)
Answer:
Given Data: L1 = 1m , R1 = 6 Ohm, L2 = 70 cm = 0.7 m, R2 =
In the usual notation,
R = p × L / A
R1 = p × L1 / A And R2= p × L1 / A
R2 / R1 = L2 / L1
= 0.7/ 1 = 0.7
R2 = 0.7 R1 = 0.7 × 6 Ohm
= 4.2 Ohm
This is
the required resistance.
B. When two resistors are connected in series, their effective resistance is 80 ohm. When they are connected in parallel, their effective resistance is 20 Ohm. What are the values of the two resistances?
Answer:
Given Data: Rs = 80 Ohm, Rp = 20 Ohm
R1 = ? R2 = ?
Series combination : Rs = R1 + R2 = 80 Ohm …… (1)
Parallel combination : 1/Rp = 1/R1 + 1/R2 = 1/20 Ohm ……….. (2)
Combining
equation (1) and (2) and disgrading the unit of resistance, we have,
C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current?
Answer:
Given Data : Q = 420 C, t = 5 minutes = 5 × 60 s = 300s, I = ?
I = Q / T
I = 420/300
I = 1.4
A
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