# SSLC Physics Solution Chapter1 Effects of Electric Current

Effects of current is one of the crucial subtopics under Potential and Current Electricity. Although it is a very early chapter, this one explains in detail the various properties and effects of electric current and is crucial to achieving high marks in SSLC Physics.

There are several different kinds of difficulties in this chapter. Each problem type has multiple problems with various choices, with the correct response eliminating all other incorrect answers. Under the heading "Effects of Electric Current," we are outlining the resolution and steps for each issue.

 Board SCERT, Kerala Text Book SCERT Based Class SSLC Subject Physics Solution Chapter Chapter 1 Chapter Name Effects of Electric Current Category Kerala SSLC

## Kerala Syllabus SSLC Class 10 Physics Textbook Solution Chapter 1 Effects of Electric Current

### Let us Assess

Qn 1.

Fuse wire is to be used by understanding the amperage correctly. Write down the amperage of the fuse wires that are currently available in the market.

• Answer)

If fuse wires of appropriate amperage are not used, excess current will flow through the circuit and it causes damage to the appliances and may result in danger.

Some of the widely used for domestic purpose fuse wires of different amperage are 1 A, 1.25 A, 1.5 A, 2.2 A, 3 A, 5 A, 10 A etc

Qn 2.

0.5 A current flows through an electric heating device connected to a 230 V supply.

1. the quantity of charge that flows through the circuit in 5 minutes is                               i. 5 C ii. 15 C iii. 150 C iv. 1500 C
2. How much is the resistance of the circuit?
3. Calculate the quantity of heat generated when current flows in the circuit for 5 minutes.
4. How much is the power of the heating device connected to the circuit if we ignore the resistance of the circuit wire?
• Answer)

I = 0.5 A, t = 5 Minutes = 5 x 300s

a)

$I = \frac{Q}{t}$

Therefore Q = I x t

= 0.5 x 300

= 150 C

b) V = 230 V , I = 0.5A

$R = \frac{V}{I}$

$= \frac{230}{0.5}$

= $460\Omega$

c) I = 0.5 A

R = $460\Omega$,

t = 5 minute = 5 x 60s

H = I2

Rt = (0.5)2 x 460 x 300

= 34500 J

d) H = 34500 J,

t = 300 s

$P =\frac{H}{t}$

$=\frac{34500}{300}$

= 115 W

Qn 3.

According to Joule’s Law, the heat generated due to the flow of current is H = IRt. Will the heat developed increase on increasing the resistance without changing the voltage? Explain.

• Answer)

Heat developed will not increase while voltage is constant, if resistance is increased, current decreases $\left [ I = \frac{V}{R} \right ]$

### Device II

Voltage = 250 V

Resistance = $2\Omega$

Time = 10 s

$I = \frac{V}{R}=\frac{250}{25}= 10 A$

H = I2Rt = 102 x 25 x 10

= 25000 J

Voltage = 250 V

Resistance = $50\Omega$

Time = 10 s

$I =\frac{V}{R}= \frac{250}{50}=5A$

H = I2Rt = 52 x 50 x 10

= 12500 J

Qn 4.

The table shows details of an electric heating device designed to work in 230 V. Complete the table by calculating the change in the heat and power on changing the voltage and resistance of the device. Analyze the table and answer the following questions.

1. How does the voltage under which a device works affect its functioning?
2. What change happens to power on increasing the resistance without changing the voltage?
3. What change is to be brought about in the construction of household heating devices in order to increase their power?
• Answer)

a) If voltage is increased without any change in resistance, power increases.

b) Resistance increases, current decreases, power also decreases

c) Decrease the resistance of the device and increase current. English Medium Tu

Qn 5.

a. Complete the table based on the amperage of the fuse wire.

b. The amperage of the fuse wire used in a circuit that works on 230 V is 2.2 A. If so the power of the device is
i. less than 300 W
ii. 300 W to 500 W
iii. between 500 Wand 510 W
iv. more than 510 W

• Answer)

Image

Between 500 W and 510 W

Qn 6.

A 230 V, 115 W filament lamp works in a circuit for 10 minutes,
a. What is the current flowing through the bulb?
b. How much is the quantity of charge that flows through the bulb in 10 minutes?

• Answer)

a) $I = \frac{P}{V} = \frac{115}{230} = 0.5 A$

b) Q = I x t = 0.5 x 600 = 300 C

Qn 7.

An electric heater conducts a 4 A current when 60 V is applied across its terminals. What will be current if the potential difference is 120 V?

• Answer)

$I = \frac{P}{V} = \frac{60}{4} = 15\Omega$

When 120 V is supplied  $I = \frac{V}{R} = \frac{120}{15} = 8A$

Qn 8.

Three resistors of 2 ? 3 ? and 6 ? are given in the class.
a. What is the highest resistance that you can get using all of them?
b. What is the least resistance that you can get using all of them?
c. Can you make a resistance 4.5? using these three? Draw the circuit.

• Answer)

a) R1 = $2\Omega$ , R$3\Omega$ ,  R3 = $6\Omega$

R1 = R1 + R2 + R$11\Omega$

b)

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$

$=\frac{3+2+1}{6}$

$=\frac{6}{6}=1$

$\therefore R = 1\Omega$

Image

Qn 9.

A girl has many resistors of 2 Q each. She needs a circuit of 9 Q resistance. For this draw a circuit with the minimum number of resistors.

• Answer)

Image

R = $9\Omega$

Qn 10.

If a bulb is lit after rejoining the parts of a broken filament, what change will occur in the intensity of the light from the lamp? What will be the change in the power of the bulb?

• Answer)

The intensity of light increases. When rejoin, the length of the filament decreases, resistance decreases current increases, power increases.

Qn 11.

Which of the following does not indicate the power of a circuit?
a. I28
b. VI
c. 1R2
d. V2/R

• Answer)

c.) 1R2

Qn 12.

How much will be the power of a 220 V, 100 W electric bulb working at 110 V?
a. 100 W
b. 75 W
c. 50 W
d. 25 W

• Answer)

$R = \frac{V^{2}}{P}= \frac{220 \times 110}{100}= 484\Omega$

When works on 110 V,

$P = \frac{V^{2}}{R}= \frac{110 \times 110}{484}= 25 W$

Qn 13.

Which of the following should be connected in parallel to a device in a circuit?
a. voltmeter
b. ammeter
c. galvanometer

• Answer)

a) voltmeter

Qn 14.

When a 12 V battery is connected to the resistor, 2.5 mA current flows through the circuit. If so what is the resistance of the resistor?

• Answer)

V = 12 V,

I = 2.5 mA

= 2.5 x 10-3

$R = \frac{V}{I}= \frac{12}{2.5\times 10^{-3}}= 4800 \Omega$

Qn 15.

If 0.2?, 0.3?, 0.4 ?, 0.5 ?, and 12? resistors are connected to a 9 V battery in parallel, what will be the current through the 12? resistor?

• Answer)

$I = \frac{V}{R} = \frac{9}{12} = 0.75 A$

When resistors are connected in parallel, each resistor gets 9V

Qn 16.

How many resistors of 176 ? should be connected in parallel to get 5A current from 220 V supply?
a. 2
b. 3
c. 6
d. 4

• Answer)

$\frac{V}{I} = R$

$\frac{220}{5} = \frac{176}{n}$

n = 4

Qn 17.

Depict a figure showing the arrangement of three resistors in a circuit to get an effective resistance of
(i) $9\Omega$
(ii) $4\Omega$

• Answer)

Image

### SSLC Physics Textbook Solution

Feel free to comment and share this article if you found it useful. Give your valuable suggestions in the comment session or contact us for any details regarding the HSE Kerala SSLC class 10 syllabus, Previous year question papers, and other study materials.

### Other Related Links

We hope the given HSE Kerala Board Syllabus SSLC class 10 Textbook Solutions Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium will help you.

If you have any query regarding the Higher Secondary Kerala SSLC class 10 SCERT syllabus, drop a comment below and we will get back to you at the earliest.

Keralanotes.com      Keralanotes.com      Keralanotes.com      Keralanotes.com      Keralanotes.com
To Top