Electricity

Ans. I = 10 A, V = 220 V

R = V/I = 220/10 = 22 ohm.

Ans. (a) It represents a battery which maintains a potential difference across the circuit element for the flow of current in the circuit.

(b) It is an ammeter which measures the current flowing in the circuit.

Ans. The potential difference between two points A and B is said to be one volt if 1 joule of work is done to move 1 coulomb of charge from one point to another point.

Ans. The connecting cord of an electric heater does not glow because its resistance is less as compared to the heating element. Hence more heat is produced in the heating element as compared to the connecting cord and it glows.

Ans. In series combination R_S = nR

For parallel R_P = R/n

R₁/R₂ = nR/(R/n) = n²

Ans. D = 0.5 m = 0.5 x 10^-3 m

ρ = 1.62 x 10^-3 Ωm

R = 10 Ω

R = ρl/A = ρl/ πr² = 4ρl/πD²

l = πRD²/4ρ

l = 3.14 x 10 x (5 x 10^-4)² / 4 x 1.62 x 10^-3 = 121.14 m

Length of the wire, l = 121.14 m

R = ρl/A = ρl/ πr² = 4ρl/πD²

R’ = ρl/A = ρl/ πr² = ρl/ π(2D/2)² = 1/4 x 4ρl/πD²

R’ = R/4

Ans. Alloys are used in electricity heating devices rather than pure metal because resistivity of an alloy is more and hence more heat is produced in any alloy. Moreover, alloy does not burn (or oxidize easily at higher temperature).

Ans. Resistance depends on the following factors:

(i) R ∝ l (length of the conductor)

(ii) R ∝ 1/A (area of cross-section)

(iii) R ∝ t (temperature)

Ans. Charge on 1 electron = 1.6 x 10^-19 C

1 electron = 1.6 x 10^-19 C

1 C = 1/(1.6 x 10¹⁹)

1 C = (1/1.6) x 10^-19 electrons

1 C = 10/16 x 10^-19 electrons

1 C = 6.25 x 10¹⁸ electrons

Ans. An electric circuit is a continuous and closed path of an electric current. If the electric circuit complete, current can flow through the circuit.

Ans. SI unit of electric current is Ampere. Current is said to be 1 ampere, if 1 coulomb charge flows per second across a cross-section of conductor.

Ans. Ammeter consists of a wire of low resistance when connected in parallel, a large amount of current passes through it hence gets burnt i.e., short circuited.

Ans. When the resistance are joined in parallel, the resultant resistance in parallel arrangements is given by

1/R = 1/R₁ + 1/R₂ + 1/R₃ + ….

(a) 1/R = 1/1 + 1/10⁶

R = 1 Ω

(b) 1/1 + 1/10³ + 1/10⁶

R = 1 Ω

Ans. resistance of electric iron (R) = 20 Ω, current (I) = 5 A and time = 30 sec

Heat generated (H) = I²Rt = 5² x 20 x 30 = 15000 Joules.

Ans. Charge transferred (Q) = 96000 C

Time = 1 hour = 60 x 60 = 3600 sec

Potential difference (V) = 50 V

Heat generated (H) = VIt = V.Q = 50 x 96000 = 4800000 Joules = 4.8 x 10⁶ Joules

Ans. It is given that current drawn by electric motor (I) = 5 A

Line voltage V = 220V

Time = 2 hours

Power of motor (P) = VI = 220 x 5 = 1100 W and the energy consumed E = Pt x 1100 x 2 h = 2200 Wh or 2.2 KWh

Ans. A voltmeter is always connected in parallel to resistance across the point between which the potential difference is to be measured.

Ans. Voltage of battery = V = 12 V

Current (I) = 2.5m A = 2.5 x 10^-3 A

Resistance (R) = V/I = 12/(2.5 x 10^-3) = 4800 Ω

Ans. Current drawn by 1^st lamp rated 100 W at 220 V = P/V = 100/220 = 5/11 A

Current drawn by 2^nd lamp rated 60 W at 220 V = 60/220 = 3/11 A

In parallel arrangement the total current = 3/11 + 5/11 = 8/11 = 0.73 A

Ans. Resistance of electric heater (R) = 8 Ω

Current (I) = 15 A

Rate at which heat developed in the heater = I²Rt/t = 15 x 15 x 8 = 1800 W

Ans. V = IR

V = constant

I₁/I₂ = R/2R

I₁/I₂ = ½

The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at higher temperatures. Therefore, conductors of electric heating devices, such as toasters and electric irons, are made up of an alloy rather than pure metal.

Ans. Let three resistors R₁, R₂ and R₃ are connected in series which are also connected with a battery, an ammeter and a key as shown in figure.

When key is closed, the current starts flowing through the circuit. Take the reading of ammeter. Now change the position of ammeter to anywhere in between the resistors and take its reading. We will observe that in both the cases reading of ammeter will be same showing same current flows through every part of the circuit above.

(b) Given,
R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 15 Ω, V = 30 V
Total resistance, R = R₁ + R₂ + R₃ [∵ 5 Ω, 10 Ω and 15 Ω are connected in series]
= 5 + 10 + 15
= 30 Ω
Potential difference, V = 30 V
Current in the circuit, I = ?
From Ohm’s law.
I = V/R = 30/30 = 1 A
Current flowing in the circuit = 1 A
Potential difference across 15 Ω resistors = IR₃ = 1 × 15 = 15 V

Let R₁, R₂ and R₃ are three resistance connected in parallel to one another and R is the equivalent resistance of the circuit. A battery of V volts has been applied across the ends of this combination. When the switch of the key is closed, current I flow in the circuit such that,

From ohm’s law,

I = V/R

I₁ = V/R₁

I₂ = V/R₂

I₃ = V/R₃

I = I₁ + I₂ + I₃

Putting the values of I, I1, I2 and I3

V/R = V/R₁ + V/R₂ + V/R₃

V(1/R) = V (1/R₁ + 1/R₂ + 1/R₃)

1/R = 1/R₁ + 1/R₂ + 1/R₃

Let R_P is the equivalent resistance of resistors connected in parallel

⸫ 1/R_P = 1/20 + 1/20

1/R_P = 2/20 = 1/10

R_P = 10 Ω

Now, equivalent circuit becomes

10 Ω and 10 Ω are connected in series.
Equivalent resistance of the circuit = 10 Ω + 10 Ω = 20 Ω

The resistance of the heating element of an electric oven is very high. As the current flows through the heating element it becomes too hot and glows red. On the other hand, the cord of an electric oven low resistance hence it does not become red during the flow of current.

Galvanometer is a device that detects the presence of current in a circuit. It is also used for measuring the amount of current in the circuit.

It is observed that total current I is equal to the sum of separate currents.
I = I₁ + I₂ + I₃ …(i)
Let Rp be the equivalent resistance of the parallel combination of resistors.
Hence, if n resistors are connected in parallel, then the equivalent resistance of the circuit is given by –

Given, two resistors of 12 Ω connected in parallel

V = 6 V

1/R_eq = 1/R₁ + 1/R₂

1/R_eq = 1/12 + 1/12

            = 2/12

R_eq = 6 Ω

According to ohm’s law V = IR

6 = I x 6

I = 6/6

= 1 amp

(a) the total resistance of the circuit,
(b) the current through the circuit,
(c) the potential difference across the (i) electric lamp and (ii) conductor, and
(d) power of the lamp.

Ans. (a) Given R1 = 20 Ω, R2 = 4 Ω

Since, in series

R = R1 + R2

Total resistance of circuit:

R = 20 + 4

= 24 Ω

(b) Current through circuit:

V = 6 V, R = 24 Ω

According to ohm’s law

V = IR

So, I = V/R

I = 6/24

= 1/4 = 0.25 Ampere

(c) (i) Potential difference across conduction:

I = ¼, R = 20 Ω

V₁ = IR₁

V₁ = ¼ X 20

            = 5V

(ii) Potential difference across lamp

V₂ = IR₂

= ¼ x 4

V₂ = 1 V

(d) Power of lamp:

P = I²R

= (1/4)² x 20

= ¼ x ¼ x 20

5/4 Watt

= 1.25 watt

The straight line in the graph signify that potential difference and current are directly proportional to each other.
The method of determining resistance of resistor using the graph is by Ohm’s law,
V = IR and by calculating the slope from the points mentioned on the graph

This is called the zero error of the scale of ammeter or voltmeter. If there is a zero error then this error is subtracted from the value that depicts when the circuit is closed otherwise accurate readings will not be recorded.

Ampere is the SI unit of current.
1 Ampere current can be defined as a unit charge flowing per second in the circuit.

1 amp = 1 coulomb/1 second

Voltmeter measures the potential difference across two points in a circuit. It is always connected in parallel in the circuit.